description: Mocha wants to be an astrologer. There are n stars which can be seen in Zhijiang, and the brightness of the i-th star is a_i. 
  
  Mocha considers that these n stars form a constellation, and she uses (a_1,a_2,…,a_n) to show its state. A state is called mathematical if all of the following three conditions are satisfied:
  
    * For all i (1≤ i≤ n), a_i is an integer in the range [l_i, r_i].
    * ∑  _{i=1} ^ n a_i ≤ m.
    * \gcd(a_1,a_2,…,a_n)=1.
  
  
  
  Here, \gcd(a_1,a_2,…,a_n) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers a_1,a_2,…,a_n.
  
  Mocha is wondering how many different mathematical states of this constellation exist. Because the answer may be large, you must find it modulo 998 244 353.
  
  Two states (a_1,a_2,…,a_n) and (b_1,b_2,…,b_n) are considered different if there exists i (1≤ i≤ n) such that a_i ≠ b_i.
  
  Input
  
  The first line contains two integers n and m (2 ≤ n ≤ 50, 1 ≤ m ≤ 10^5) — the number of stars and the upper bound of the sum of the brightness of stars.
  
  Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — the range of the brightness of the i-th star.
  
  Output
  
  Print a single integer — the number of different mathematical states of this constellation, modulo 998 244 353.
  
  Examples
  
  Input
  
  
  2 4
  1 3
  1 2
  
  
  Output
  
  
  4
  
  Input
  
  
  5 10
  1 10
  1 10
  1 10
  1 10
  1 10
  
  
  Output
  
  
  251
  
  Input
  
  
  5 100
  1 94
  1 96
  1 91
  4 96
  6 97
  
  
  Output
  
  
  47464146
  
  Note
  
  In the first example, there are 4 different mathematical states of this constellation:
  
    * a_1=1, a_2=1.
    * a_1=1, a_2=2.
    * a_1=2, a_2=1.
    * a_1=3, a_2=1.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int MAXN = 5e1 + 10, MOD = 998244353;
  const int MAXM = 1e5 + 10;
  template <int MOD_>
  struct modnum {
    static constexpr int MOD = MOD_;
    static_assert(MOD_ > 0, "MOD must be positive");
  
   private:
    using long long = long long;
    int v;
    static int minv(int a, int m) {
      a %= m;
      assert(a);
      return a == 1 ? 1 : int(m - long long(minv(m, a)) * long long(m) / a);
    }
  
   public:
    modnum() : v(0) {}
    modnum(long long v_) : v(int(v_ % MOD)) {
      if (v < 0) v += MOD;
    }
    explicit operator int() const { return v; }
    friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
      return out << int(n);
    }
    friend std::istream& operator>>(std::istream& in, modnum& n) {
      long long v_;
      in >> v_;
      n = modnum(v_);
      return in;
    }
    friend bool operator==(const modnum& a, const modnum& b) {
      return a.v == b.v;
    }
    friend bool operator!=(const modnum& a, const modnum& b) {
      return a.v != b.v;
    }
    modnum inv() const {
      modnum res;
      res.v = minv(v, MOD);
      return res;
    }
    friend modnum inv(const modnum& m) { return m.inv(); }
    modnum neg() const {
      modnum res;
      res.v = v ? MOD - v : 0;
      return res;
    }
    friend modnum neg(const modnum& m) { return m.neg(); }
    modnum operator-() const { return neg(); }
    modnum operator+() const { return modnum(*this); }
    modnum& operator++() {
      v++;
      if (v == MOD) v = 0;
      return *this;
    }
    modnum& operator--() {
      if (v == 0) v = MOD;
      v--;
      return *this;
    }
    modnum& operator+=(const modnum& o) {
      v -= MOD - o.v;
      v = (v < 0) ? v + MOD : v;
      return *this;
    }
    modnum& operator-=(const modnum& o) {
      v -= o.v;
      v = (v < 0) ? v + MOD : v;
      return *this;
    }
    modnum& operator*=(const modnum& o) {
      v = int(long long(v) * long long(o.v) % MOD);
      return *this;
    }
    modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
    friend modnum operator++(modnum& a, int) {
      modnum r = a;
      ++a;
      return r;
    }
    friend modnum operator--(modnum& a, int) {
      modnum r = a;
      --a;
      return r;
    }
    friend modnum operator+(const modnum& a, const modnum& b) {
      return modnum(a) += b;
    }
    friend modnum operator-(const modnum& a, const modnum& b) {
      return modnum(a) -= b;
    }
    friend modnum operator*(const modnum& a, const modnum& b) {
      return modnum(a) *= b;
    }
    friend modnum operator/(const modnum& a, const modnum& b) {
      return modnum(a) /= b;
    }
  };
  using num = modnum<MOD>;
  const int MAXV = 2e6 + 10;
  int pf[MAXV];
  void prep() {
    memset(pf, -1, sizeof(pf));
    for (int i = 2; i < MAXV; i++)
      if (pf[i] == -1) {
        for (int j = i; j < MAXV; j += i) pf[j] = i;
      }
  }
  int get_mul(int c) {
    int ans = 0;
    while (c > 1) {
      int p = pf[c];
      c /= p;
      if (c % p == 0) return 0;
      ans ^= 1;
    }
    return (ans == 0 ? 1 : -1);
  }
  int n, m;
  array<int, 2> a[MAXN], b[MAXN];
  num get(int d) {
    int bound = m / d;
    for (int i = 0; i < n; i++) {
      b[i][0] = (a[i][0] + d - 1) / d;
      b[i][1] = a[i][1] / d;
    }
    vector<num> pre(bound + 1, 1), cur(bound + 1);
    for (int rep = 0; rep < n; rep++) {
      int l = b[rep][0], r = b[rep][1];
      auto get = [&](int i) -> num { return i >= 0 ? pre[i] : 0; };
      for (int i = 0; i <= bound; i++) {
        cur[i] = get(i - l) - get(i - r - 1);
      }
      for (int i = 0; i <= bound; i++) {
        pre[i] = cur[i], cur[i] = 0;
        if (i) pre[i] += pre[i - 1];
      }
    }
    return pre[bound];
  }
  void solve() {
    prep();
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> a[i][0] >> a[i][1];
    num ans = 0;
    for (int d = 1; d <= m; d++) {
      int mul = get_mul(d);
      ans += get(d) * mul;
    }
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T = 1;
    while (T--) solve();
  }
  