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>
  using namespace std;
  char _c;
  bool _f;
  template <class T>
  inline void IN(T &x) {
    x = 0, _f = 0;
    while (_c = getchar(), !isdigit(_c))
      if (_c == '-') _f = 1;
    while (isdigit(_c)) x = x * 10 + _c - '0', _c = getchar();
    if (_f) x = -x;
  }
  const int mod = 998244353;
  inline int mul(int x, int y) { return 1ll * x * y % mod; }
  inline void sub(int &x, int y) {
    x -= y;
    if (x < 0) x += mod;
  }
  inline void pls(int &x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  inline int dec(int x, int y) {
    x -= y;
    if (x < 0) x += mod;
    return x;
  }
  inline int add(int x, int y) {
    x += y;
    if (x >= mod) x -= mod;
    return x;
  }
  inline int modpow(int x, long long y, int res = 1) {
    for (y = (y + mod - 1) % (mod - 1); y; y >>= 1, x = mul(x, x))
      if (y & 1) res = mul(x, res);
    return res;
  }
  template <class T>
  inline void chkmin(T &x, T y) {
    if (x > y) x = y;
  }
  template <class T>
  inline void chkmax(T &x, T y) {
    if (x < y) x = y;
  }
  const int LogN = 20;
  int lim, tim, inv[LogN];
  vector<int> w[LogN], rev[LogN];
  inline void init_w(int len = 1 << 19) {
    w[19].resize(len), w[19][0] = 1, w[19][1] = modpow(3, (mod - 1) >> 20);
    for (int i = 2; i < len; ++i) w[19][i] = mul(w[19][i - 1], w[19][1]);
    for (int i = 18; i >= 0; --i) {
      w[i].resize(len >>= 1);
      for (int j = 0; j < len; ++j) w[i][j] = w[i + 1][j << 1];
    }
  }
  inline void init_r(int len) {
    for (lim = 1, tim = 0; lim < len; lim <<= 1, ++tim)
      ;
    if (rev[tim].size()) return;
    rev[tim].resize(lim), inv[tim] = modpow(lim, -1);
    for (int i = 0; i < lim; ++i)
      rev[tim][i] = (rev[tim][i >> 1] >> 1) | ((i & 1) << (tim - 1));
  }
  inline void ntt(vector<int> &f, short typ) {
    static unsigned long long g[1 << 20 | 5];
    for (int i = 0; i < lim; ++i) g[rev[tim][i]] = f[i];
    for (int p = 1, s = 0, t = 0; p < lim; p <<= 1, ++t)
      for (int k = 0; k < lim; k += p << 1)
        for (int l = k; l < k + p; ++l)
          s = mul(g[l + p] % mod, w[t][l - k]), g[l + p] = g[l] + mod - s,
          g[l] += s;
    for (int i = 0; i < lim; ++i) f[i] = g[i] % mod;
    if (~typ) return;
    std ::reverse(++f.begin(), f.end());
    for (int i = 0; i < lim; ++i) f[i] = mul(f[i], inv[tim]);
  }
  const int N = 1e5 + 5;
  const int inv2 = modpow(2, -1);
  int n, m, a[N], b[N], fac[N];
  vector<int> f[N << 2][4], g[2][4];
  void solve(int x, int l, int r, bool typ) {
    const int len = r - l + 2;
    for (int o = 0; o < 4; ++o) f[x][o].resize(len);
    if (l == r) return f[x][(b[l] == 1) * 3][1] = 1 + (b[l] != 1), void();
    const int mid = l + r >> 1, lc = x << 1, rc = x << 1 | 1;
    solve(lc, l, mid, 0), solve(rc, mid + 1, r, 1);
    const int ln = mid - l + 2, rn = r - mid + 1;
    for (int o0 = 0; o0 < 4; ++o0)
      for (int o1 = 0; o1 < 4; ++o1) {
        int t0 = (o0 & 1) | (o1 & 2), t2 = (o0 & 2) | (o1 & 1),
            val = (t2 == 3) ? 2 : (t2 == 0 ? inv2 : 1);
        pls(f[x][t0][2], mul(f[lc][o0][1], f[rc][o1][1]));
        pls(f[x][0][1], mul(mul(f[lc][o0][1], f[rc][o1][1]), val));
        for (int i = 3; i <= ln; ++i)
          pls(f[x][t0][i], mul(f[lc][o0][i - 1], f[rc][o1][1]));
        for (int j = 3; j <= rn; ++j)
          pls(f[x][t0][j], mul(f[lc][o0][1], f[rc][o1][j - 1]));
        for (int i = 2; i < ln; ++i)
          pls(f[x][o0 & 1][i], mul(mul(f[lc][o0][i], f[rc][o1][1]), val));
        for (int j = 2; j < rn; ++j)
          pls(f[x][o1 & 2][j], mul(mul(f[lc][o0][1], f[rc][o1][j]), val));
      }
    init_r(len);
    for (int t0 = 0; t0 < 4; ++t0) g[0][t0].resize(lim), g[1][t0].resize(lim);
    for (int o0 = 0; o0 < 4; ++o0)
      f[lc][o0][0] = f[lc][o0][1] = 0, f[lc][o0].resize(lim), ntt(f[lc][o0], 1);
    for (int o1 = 0; o1 < 4; ++o1)
      f[rc][o1][0] = f[rc][o1][1] = 0, f[rc][o1].resize(lim), ntt(f[rc][o1], 1);
    for (int o0 = 0; o0 < 4; ++o0)
      for (int o1 = 0; o1 < 4; ++o1) {
        int t0 = (o0 & 1) | (o1 & 2), t2 = (o0 & 2) | (o1 & 1),
            val = (t2 == 3) ? 2 : (t2 == 0 ? inv2 : 1);
        for (int i = 0; i < lim; ++i)
          g[0][t0][i] = mul(f[lc][o0][i], f[rc][o1][i]);
        ntt(g[0][t0], -1);
        for (int i = 4; i < len; ++i) pls(f[x][t0][i], g[0][t0][i]);
        for (int i = 4; i < len; ++i) pls(f[x][t0][i - 1], mul(g[0][t0][i], val));
      }
  }
  int main() {
    init_w();
    IN(n);
    for (int i = 1; i <= n; ++i) IN(a[i]);
    fac[0] = 1;
    for (int i = 1; i <= n; ++i) fac[i] = mul(fac[i - 1], i);
    bool flag = false;
    for (int l = 1, r; l <= n; l = r + 1) {
      if (r = l + a[l] - 1, r > n) {
        flag = true;
        break;
      }
      for (int i = l; i <= r; ++i)
        if (a[i] != a[l]) flag = true;
      b[++m] = a[l];
    }
    if (flag) return puts("0"), 0;
    solve(1, 1, m, 0);
    int ans = 0;
    for (int i = 1; i <= m; ++i)
      for (int o = 0; o < 4; ++o)
        pls(ans, mul(f[1][o][i], mul(fac[i], ((m - i) & 1) ? mod - 1 : 1)));
    printf("%d\n", ans);
    return 0;
  }
  