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>
  const int maxn = 55, N = 1e5, mod = 998244353;
  int n, m, l[maxn], r[maxn], vis[N + 5], pr[N + 5], mu[N + 5], prcnt;
  void Sieve(int N) {
    mu[1] = 1;
    for (int i = 2; i <= N; ++i) {
      if (!vis[i]) pr[++prcnt] = i, mu[i] = -1;
      for (int j = 1; j <= prcnt && i * pr[j] <= N; ++j) {
        vis[i * pr[j]] = 1;
        if (i % pr[j] == 0) {
          mu[i * pr[j]] = 0;
          break;
        }
        mu[i * pr[j]] = -mu[i];
      }
    }
  }
  int f[N + 5], s[N + 5];
  int Solve(int d) {
    f[0] = s[0] = 1;
    for (int i = 1; i <= m / d; ++i) f[i] = 0, s[i] = 1;
    for (int i = 1; i <= n; ++i) {
      int L = (l[i] - 1) / d + 1, R = r[i] / d;
      if (L > R) return 0;
      for (int j = 0; j < L; ++j) f[j] = 0;
      for (int j = L; j <= R; ++j) f[j] = s[j - L];
      for (int j = R + 1; j <= m / d; ++j)
        f[j] = (s[j - L] - s[j - R - 1] + mod) % mod;
      s[0] = f[0];
      for (int j = 1; j <= m / d; ++j) s[j] = (s[j - 1] + f[j]) % mod;
    }
    return s[m / d];
  }
  int main() {
    scanf("%d %d", &n, &m), Sieve(m);
    for (int i = 1; i <= n; ++i) scanf("%d %d", &l[i], &r[i]);
    int ans = 0;
    for (int i = 1; i <= m; ++i)
      mu[i] == 1 ? ans = (ans + Solve(i)) % mod
                 : (mu[i] == -1 ? ans = (ans - Solve(i) + mod) % mod : 0);
    return printf("%d", ans), 0;
  }
  