description: For a permutation p of numbers 1 through n, we define a stair array a as follows: a_i is length of the longest segment of permutation which contains position i and is made of consecutive values in sorted order: [x, x+1, …, y-1, y] or [y, y-1, …, x+1, x] for some x ≤ y. For example, for permutation p = [4, 1, 2, 3, 7, 6, 5] we have a = [1, 3, 3, 3, 3, 3, 3]. 
  
  You are given the stair array a. Your task is to calculate the number of permutations which have stair array equal to a. Since the number can be big, compute it modulo 998 244 353. Note that this number can be equal to zero.
  
  Input
  
  The first line of input contains integer n (1 ≤ n ≤ 10^5) — the length of a stair array a.
  
  The second line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n).
  
  Output
  
  Print the number of permutations which have stair array equal to a. Since the number can be big, compute it modulo 998 244 353.
  
  Examples
  
  Input
  
  
  6
  3 3 3 1 1 1
  
  
  Output
  
  
  6
  
  
  Input
  
  
  7
  4 4 4 4 3 3 3
  
  
  Output
  
  
  6
  
  
  Input
  
  
  1
  1
  
  
  Output
  
  
  1
  
  
  Input
  
  
  8
  2 2 2 2 2 2 1 1
  
  
  Output
  
  
  370
  
  
  Input
  
  
  4
  3 2 3 1
  
  
  Output
  
  
  0
solution: #include <bits/stdc++.h>
  template <typename T>
  bool chkmax(T &x, T y) {
    return x < y ? x = y, true : false;
  }
  template <typename T>
  bool chkmin(T &x, T y) {
    return x > y ? x = y, true : false;
  }
  using namespace std;
  long long ksm(long long a, long long b) {
    if (!b) return 1;
    long long ns = ksm(a, b >> 1);
    ns = ns * ns % 998244353;
    if (b & 1) ns = ns * a % 998244353;
    return ns;
  }
  const int md = 998244353;
  inline void add(int &x, int y) {
    x += y;
    if (x >= md) {
      x -= md;
    }
  }
  inline void sub(int &x, int y) {
    x -= y;
    if (x < 0) {
      x += md;
    }
  }
  inline int mul(int x, int y) { return (long long)x * y % md; }
  inline int power(int x, int y) {
    int res = 1;
    for (; y; y >>= 1, x = mul(x, x)) {
      if (y & 1) {
        res = mul(res, x);
      }
    }
    return res;
  }
  inline int inv(int a) {
    a %= md;
    if (a < 0) {
      a += md;
    }
    int b = md, u = 0, v = 1;
    while (a) {
      int t = b / a;
      b -= t * a;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    if (u < 0) {
      u += md;
    }
    return u;
  }
  namespace ntt {
  int base = 1, root = -1, max_base = -1;
  vector<int> rev = {0, 1}, roots = {0, 1};
  void init() {
    int temp = md - 1;
    max_base = 0;
    while (temp % 2 == 0) {
      temp >>= 1;
      ++max_base;
    }
    root = 2;
    while (true) {
      if (power(root, 1 << max_base) == 1 &&
          power(root, 1 << max_base - 1) != 1) {
        break;
      }
      ++root;
    }
  }
  void ensure_base(int nbase) {
    if (max_base == -1) {
      init();
    }
    if (nbase <= base) {
      return;
    }
    assert(nbase <= max_base);
    rev.resize(1 << nbase);
    for (int i = 0; i < 1 << nbase; ++i) {
      rev[i] = rev[i >> 1] >> 1 | (i & 1) << nbase - 1;
    }
    roots.resize(1 << nbase);
    while (base < nbase) {
      int z = power(root, 1 << max_base - 1 - base);
      for (int i = 1 << base - 1; i < 1 << base; ++i) {
        roots[i << 1] = roots[i];
        roots[i << 1 | 1] = mul(roots[i], z);
      }
      ++base;
    }
  }
  void dft(vector<int> &a) {
    int n = a.size(), zeros = __builtin_ctz(n);
    ensure_base(zeros);
    int shift = base - zeros;
    for (int i = 0; i < n; ++i) {
      if (i < rev[i] >> shift) {
        swap(a[i], a[rev[i] >> shift]);
      }
    }
    for (int i = 1; i < n; i <<= 1) {
      for (int j = 0; j < n; j += i << 1) {
        for (int k = 0; k < i; ++k) {
          int x = a[j + k], y = mul(a[j + k + i], roots[i + k]);
          a[j + k] = (x + y) % md;
          a[j + k + i] = (x + md - y) % md;
        }
      }
    }
  }
  vector<int> multiply(vector<int> a, vector<int> b) {
    int need = a.size() + b.size() - 1, nbase = 0;
    while (1 << nbase < need) {
      ++nbase;
    }
    ensure_base(nbase);
    int sz = 1 << nbase;
    a.resize(sz);
    b.resize(sz);
    bool equal = a == b;
    dft(a);
    if (equal) {
      b = a;
    } else {
      dft(b);
    }
    int inv_sz = inv(sz);
    for (int i = 0; i < sz; ++i) {
      a[i] = mul(mul(a[i], b[i]), inv_sz);
    }
    reverse(a.begin() + 1, a.end());
    dft(a);
    a.resize(need);
    return a;
  }
  vector<int> inverse(vector<int> a) {
    int n = a.size(), m = n + 1 >> 1;
    if (n == 1) {
      return vector<int>(1, inv(a[0]));
    } else {
      vector<int> b = inverse(vector<int>(a.begin(), a.begin() + m));
      int need = n << 1, nbase = 0;
      while (1 << nbase < need) {
        ++nbase;
      }
      ensure_base(nbase);
      int sz = 1 << nbase;
      a.resize(sz);
      b.resize(sz);
      dft(a);
      dft(b);
      int inv_sz = inv(sz);
      for (int i = 0; i < sz; ++i) {
        a[i] = mul(mul(md + 2 - mul(a[i], b[i]), b[i]), inv_sz);
      }
      reverse(a.begin() + 1, a.end());
      dft(a);
      a.resize(n);
      return a;
    }
  }
  }  // namespace ntt
  using ntt::inverse;
  using ntt::multiply;
  vector<int> &operator+=(vector<int> &a, const vector<int> &b) {
    if (a.size() < b.size()) {
      a.resize(b.size());
    }
    for (int i = 0; i < b.size(); ++i) {
      add(a[i], b[i]);
    }
    return a;
  }
  vector<int> operator+(const vector<int> &a, const vector<int> &b) {
    vector<int> c = a;
    return c += b;
  }
  vector<int> &operator-=(vector<int> &a, const vector<int> &b) {
    if (a.size() < b.size()) {
      a.resize(b.size());
    }
    for (int i = 0; i < b.size(); ++i) {
      sub(a[i], b[i]);
    }
    return a;
  }
  vector<int> operator-(const vector<int> &a, const vector<int> &b) {
    vector<int> c = a;
    return c -= b;
  }
  vector<int> &operator*=(vector<int> &a, const vector<int> &b) {
    if (min(a.size(), b.size()) < 128) {
      vector<int> c = a;
      a.assign(a.size() + b.size() - 1, 0);
      for (int i = 0; i < c.size(); ++i) {
        for (int j = 0; j < b.size(); ++j) {
          add(a[i + j], mul(c[i], b[j]));
        }
      }
    } else {
      a = multiply(a, b);
    }
    return a;
  }
  vector<int> operator*(const vector<int> &a, const vector<int> &b) {
    vector<int> c = a;
    return c *= b;
  }
  vector<int> &operator/=(vector<int> &a, const vector<int> &b) {
    int n = a.size(), m = b.size();
    if (n < m) {
      a.clear();
    } else {
      vector<int> c = b;
      reverse(a.begin(), a.end());
      reverse(c.begin(), c.end());
      c.resize(n - m + 1);
      a *= inverse(c);
      a.erase(a.begin() + n - m + 1, a.end());
      reverse(a.begin(), a.end());
    }
    return a;
  }
  vector<int> operator/(const vector<int> &a, const vector<int> &b) {
    vector<int> c = a;
    return c /= b;
  }
  vector<int> &operator%=(vector<int> &a, const vector<int> &b) {
    int n = a.size(), m = b.size();
    if (n >= m) {
      vector<int> c = (a / b) * b;
      a.resize(m - 1);
      for (int i = 0; i < m - 1; ++i) {
        sub(a[i], c[i]);
      }
    }
    return a;
  }
  vector<int> operator%(const vector<int> &a, const vector<int> &b) {
    vector<int> c = a;
    return c %= b;
  }
  vector<int> derivative(const vector<int> &a) {
    int n = a.size();
    vector<int> b(n - 1);
    for (int i = 1; i < n; ++i) {
      b[i - 1] = mul(a[i], i);
    }
    return b;
  }
  vector<int> primitive(const vector<int> &a) {
    int n = a.size();
    vector<int> b(n + 1), invs(n + 1);
    for (int i = 1; i <= n; ++i) {
      invs[i] = i == 1 ? 1 : mul(md - md / i, invs[md % i]);
      b[i] = mul(a[i - 1], invs[i]);
    }
    return b;
  }
  vector<int> logarithm(const vector<int> &a) {
    vector<int> b = primitive(derivative(a) * inverse(a));
    b.resize(a.size());
    return b;
  }
  vector<int> exponent(const vector<int> &a) {
    vector<int> b(1, 1);
    while (b.size() < a.size()) {
      vector<int> c(a.begin(), a.begin() + min(a.size(), b.size() << 1));
      add(c[0], 1);
      vector<int> old_b = b;
      b.resize(b.size() << 1);
      c -= logarithm(b);
      c *= old_b;
      for (int i = b.size() >> 1; i < b.size(); ++i) {
        b[i] = c[i];
      }
    }
    b.resize(a.size());
    return b;
  }
  vector<int> power(const vector<int> &a, int m) {
    int n = a.size(), p = -1;
    vector<int> b(n);
    for (int i = 0; i < n; ++i) {
      if (a[i]) {
        p = i;
        break;
      }
    }
    if (p == -1) {
      b[0] = !m;
      return b;
    }
    if ((long long)m * p >= n) {
      return b;
    }
    int mu = power(a[p], m), di = inv(a[p]);
    vector<int> c(n - m * p);
    for (int i = 0; i < n - m * p; ++i) {
      c[i] = mul(a[i + p], di);
    }
    c = logarithm(c);
    for (int i = 0; i < n - m * p; ++i) {
      c[i] = mul(c[i], m);
    }
    c = exponent(c);
    for (int i = 0; i < n - m * p; ++i) {
      b[i + m * p] = mul(c[i], mu);
    }
    return b;
  }
  vector<int> sqrt(const vector<int> &a) {
    vector<int> b(1, 1);
    while (b.size() < a.size()) {
      vector<int> c(a.begin(), a.begin() + min(a.size(), b.size() << 1));
      vector<int> old_b = b;
      b.resize(b.size() << 1);
      c *= inverse(b);
      for (int i = b.size() >> 1; i < b.size(); ++i) {
        b[i] = mul(c[i], md + 1 >> 1);
      }
    }
    b.resize(a.size());
    return b;
  }
  vector<int> multiply_all(int l, int r, vector<vector<int>> &all) {
    if (l > r) {
      return vector<int>();
    } else if (l == r) {
      return all[l];
    } else {
      int y = l + r >> 1;
      return multiply_all(l, y, all) * multiply_all(y + 1, r, all);
    }
  }
  vector<int> evaluate(const vector<int> &f, const vector<int> &x) {
    int n = x.size();
    if (!n) {
      return vector<int>();
    }
    vector<vector<int>> up(n * 2);
    for (int i = 0; i < n; ++i) {
      up[i + n] = vector<int>{(md - x[i]) % md, 1};
    }
    for (int i = n - 1; i; --i) {
      up[i] = up[i << 1] * up[i << 1 | 1];
    }
    vector<vector<int>> down(n * 2);
    down[1] = f % up[1];
    for (int i = 2; i < n * 2; ++i) {
      down[i] = down[i >> 1] % up[i];
    }
    vector<int> y(n);
    for (int i = 0; i < n; ++i) {
      y[i] = down[i + n][0];
    }
    return y;
  }
  vector<int> interpolate(const vector<int> &x, const vector<int> &y) {
    int n = x.size();
    vector<vector<int>> up(n * 2);
    for (int i = 0; i < n; ++i) {
      up[i + n] = vector<int>{(md - x[i]) % md, 1};
    }
    for (int i = n - 1; i; --i) {
      up[i] = up[i << 1] * up[i << 1 | 1];
    }
    vector<int> a = evaluate(derivative(up[1]), x);
    for (int i = 0; i < n; ++i) {
      a[i] = mul(y[i], inv(a[i]));
    }
    vector<vector<int>> down(n * 2);
    for (int i = 0; i < n; ++i) {
      down[i + n] = vector<int>(1, a[i]);
    }
    for (int i = n - 1; i; --i) {
      down[i] = down[i << 1] * up[i << 1 | 1] + down[i << 1 | 1] * up[i << 1];
    }
    return down[1];
  }
  int n, m, a[600005];
  vector<int> v;
  vector<vector<vector<int>>> solve(int l, int r) {
    if (l == r) {
      return {
          {{1, 2}, (v[l] == 1) ? vector<int>{0, 998244353 - 1} : vector<int>{0}},
          {{0, 2}, (v[l] == 1) ? vector<int>{0, 998244353 - 1} : vector<int>{0}}};
    } else {
      int mid = (l + r) >> 1;
      auto L = solve(l, mid), R = solve(mid + 1, r);
      vector<vector<vector<int>>> M(2, vector<vector<int>>(2));
      for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++)
          for (int k = 0; k < 2; k++) M[i][j] = M[i][j] + L[i][k] * R[k][j];
      return M;
    }
  }
  int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%d", a + i);
    }
    int c = 0, len = 0;
    for (int i = 0; i < n; i++) {
      if (len == 0) {
        c = a[i];
        len = a[i];
        v.push_back(a[i]);
      }
      if (a[i] != c) {
        puts("0");
        return 0;
      }
      len--;
    }
    if (len != 0) {
      puts("0");
      return 0;
    }
    m = v.size();
    auto Mans = solve(0, m - 1);
    auto ans = Mans[1][0] + Mans[1][1];
    long long w = (m % 2 == 0) ? 1 : 998244353 - 1, d = 0;
    for (int i = 1; i < ans.size(); i++) {
      w = w * (998244353 - i) % 998244353;
      d = (d + w * ans[i]) % 998244353;
    }
    printf("%lld\n", d);
  }
  