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 <typename T, typename U>
  bool ckmin(T &a, const U &b) {
    return b < a ? a = b, true : false;
  }
  template <typename T, typename U>
  bool ckmax(T &a, const U &b) {
    return b > a ? a = b, true : false;
  }
  const int N = 1e5 + 5;
  const long long MOD = 998244353;
  int n, m;
  vector<int> mobius, lpf;
  vector<array<int, 2>> ranges;
  void least_prime_factor() {
    for (int i = 2; i < N; i++)
      if (!lpf[i])
        for (int j = i; j < N; j += i)
          if (!lpf[j]) lpf[j] = i;
  }
  void Mobius() {
    for (int i = 1; i < N; i++) {
      if (i == 1)
        mobius[i] = 1;
      else {
        if (lpf[i / lpf[i]] == lpf[i])
          mobius[i] = 0;
        else
          mobius[i] = -1 * mobius[i / lpf[i]];
      }
    }
  }
  long long solve(vector<array<int, 2>> &v, int sum) {
    vector<long long> cur(sum + 1, 0), nxt(sum + 1, 0);
    for (int i = v[0][0]; i <= v[0][1]; ++i) cur[i] = 1;
    for (int j = 1; j < n; ++j) {
      vector<long long> pc(sum + 2, 0);
      for (int i = 0; i < sum + 1; ++i) {
        pc[i + 1] = pc[i] + cur[i];
        pc[i + 1] %= MOD;
      }
      auto first = [&](int l, int r) -> long long {
        ckmax(l, 0);
        ckmax(r, 0);
        ckmin(l, sum);
        ckmin(r, sum);
        return pc[r + 1] - pc[l] + (pc[l] > pc[r + 1]) * MOD;
      };
      for (int i = 0; i <= sum; ++i) {
        int l = i - v[j][1], r = i - v[j][0];
        nxt[i] = first(l, r);
      }
      cur.swap(nxt);
      nxt.clear();
      pc.clear();
      nxt.assign(sum + 1, 0);
    }
    long long res = 0;
    for (auto i : cur) res += i;
    return res % MOD;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> m;
    ranges.assign(n, array<int, 2>());
    for (int i = 0; i < n; ++i) {
      cin >> ranges[i][0] >> ranges[i][1];
    }
    long long res = 0;
    lpf.assign(N, 0);
    least_prime_factor();
    mobius.assign(N, 0);
    Mobius();
    for (int g = 1; g <= m / n; ++g) {
      int cur_m = m / g;
      vector<array<int, 2>> cur_ranges = ranges;
      for (auto &i : cur_ranges) {
        i[0] = (i[0] - 1) / g + 1;
        i[1] /= g;
        if (i[1] < i[0]) goto nxt;
      }
      res += mobius[g] * solve(cur_ranges, cur_m);
      res %= MOD;
    nxt:;
    }
    cout << ((res % MOD) + MOD) % MOD;
    return 0;
  }
  