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 = 1e5 + 10, P = 998244353;
  int n, m, a[N], L[N], R[N], mu[N], f[N], g[N];
  int main() {
    mu[1] = 1;
    for (int i = 1; i < N; ++i)
      for (int j = 2 * i; j < N; j += i) mu[j] -= mu[i];
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &L[i], &R[i]);
    }
    int ans = 0;
    for (int i = 1; i < N; ++i) {
      if (mu[i]) {
        f[0] = 1;
        int mx = m / i;
        for (int z = 1; z <= mx; ++z) f[z] = 0;
        for (int k = 1; k <= n; ++k) {
          int l = (L[k] + i - 1) / i, r = R[k] / i;
          for (int z = 0; z <= mx; ++z) g[z] = 0;
          for (int z = 0; z <= mx; ++z) {
            if (l + z <= mx) g[l + z] = (g[l + z] + f[z]) % P;
            if (r + z <= mx) g[r + z + 1] = (g[r + z + 1] - f[z]) % P;
          }
          f[0] = g[0];
          for (int z = 1; z <= mx; ++z) f[z] = (f[z - 1] + g[z]) % P;
        }
        for (int z = 0; z <= mx; ++z) ans = (ans + mu[i] * f[z]) % P;
      }
    }
    if (ans < 0) ans += P;
    printf("%d\n", ans);
  }
  