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;
  template <int MOD_>
  struct modnum {
    static constexpr int MOD = MOD_;
    static_assert(MOD_ > 0, "MOD must be positive");
  
   private:
    using ll = long long;
    int v;
    static int minv(int a, int m) {
      a %= m;
      assert(a);
      return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
    }
  
   public:
    modnum() : v(0) {}
    modnum(ll 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) {
      ll 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 += o.v;
      if (v >= MOD) v -= MOD;
      return *this;
    }
    modnum& operator-=(const modnum& o) {
      v -= o.v;
      if (v < 0) v += MOD;
      return *this;
    }
    modnum& operator*=(const modnum& o) {
      v = int(ll(v) * ll(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;
    }
  };
  template <typename T>
  T pow(T a, long long b) {
    assert(b >= 0);
    T r = 1;
    while (b) {
      if (b & 1) r *= a;
      b >>= 1;
      a *= a;
    }
    return r;
  }
  using num = modnum<998244353>;
  const int X = 2.1e5;
  bitset<X> is_prime;
  vector<int> pr;
  vector<int> mu(X, 0);
  void init() {
    is_prime.flip();
    is_prime[0] = is_prime[1] = false;
    mu[1] = 1;
    for (int i = 2; i < X; i++) {
      if (is_prime[i]) {
        pr.push_back(i);
        mu[i] = -1;
      }
      for (int p : pr) {
        if (i * p >= X) break;
        is_prime[i * p] = false;
        if (i % p == 0) {
          mu[i * p] = 0;
        } else {
          mu[i * p] = -mu[i];
        }
        if (i % p == 0) break;
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    init();
    int n, m;
    cin >> n >> m;
    vector<int> l(n), r(n);
    int maxr = 0;
    for (int i = 0; i < n; i++) {
      cin >> l[i] >> r[i];
      maxr = max(maxr, r[i]);
    }
    num total = 0;
    for (int g = 1; g <= maxr; g++) {
      if (mu[g] == 0) continue;
      int S = m / g + 1;
      vector<num> dp(S, 0);
      dp[0] = 1;
      for (int i = 0; i < n; i++) {
        int newl = (l[i] + g - 1) / g;
        int newr = r[i] / g;
        vector<num> psum(S + 1, 0);
        for (int x = 0; x < S; x++) psum[x + 1] = psum[x] + dp[x];
        vector<num> ndp(S, 0);
        if (newl <= newr) {
          for (int x = 0; x < S; x++) {
            ndp[x] = psum[max(x - newl + 1, 0)] - psum[max(x - newr, 0)];
          }
        }
        dp = ndp;
      }
      num ans = 0;
      for (num v : dp) ans += v;
      total += ans * mu[g];
    }
    cout << total << '\n';
  }
  