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 N = 50, M = 1e5 + 1, MOD = 998244353;
  int l[N], r[N];
  long long f[M];
  long long g[N + 1][M];
  int main() {
    cin.tie(0), cout.tie(0);
    ios::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
    for (int i = 1; i <= m; i++) {
      int res = m;
      vector<int> cnt;
      for (int j = 0; j < n; j++) {
        int dt = (l[j] + i - 1) / i;
        res -= dt * i;
        if (dt * i > r[j])
          res = -1;
        else
          cnt.push_back((r[j] - dt * i) / i + 1);
      }
      if (res < 0)
        f[i] = 0;
      else {
        int left = res / i;
        for (int j = 0; j <= left; j++) g[0][j] = 1;
        for (int k = 0; k < n; k++) {
          g[k + 1][0] = g[k][0];
          for (int j = 0; j <= left; j++) {
            g[k + 1][j] = g[k + 1][j - 1] + g[k][j];
            if (j - cnt[k] >= 0) g[k + 1][j] -= g[k][j - cnt[k]];
            g[k + 1][j] %= MOD;
          }
        }
        f[i] = g[n][left];
      }
    }
    for (int i = m; i >= 1; i--)
      for (int j = i + i; j <= m; j += i) f[i] = (f[i] - f[j]) % MOD;
    cout << (f[1] + MOD) % MOD << "\n";
  }
  