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 + 5, mod = 998244353;
  long long f[55][N], l[N], r[N], sum[N], prim[N], nump, vis[N], mu[N];
  void init() {
    mu[1] = 1;
    for (int i = 2; i < N; i++) {
      if (!vis[i]) {
        prim[++nump] = i;
        mu[i] = -1;
      }
      for (int j = 1; j <= nump && i * prim[j] < N; j++) {
        vis[i * prim[j]] = 1;
        if (i % prim[j] == 0)
          break;
        else
          mu[i * prim[j]] = -mu[i];
      }
    }
  }
  int main() {
    long long n, m;
    scanf("%lld%lld", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%lld%lld", l + i, r + i);
    long long ans = 0;
    init();
    for (int d = 1; d <= m; d++) {
      f[0][0] = 1;
      long long cnt = 0;
      if (mu[d]) {
        for (int i = 1; i <= n; i++) {
          sum[0] = f[i - 1][0];
          for (int j = 1; j <= m / d; j++)
            sum[j] = (sum[j - 1] + f[i - 1][j]) % mod;
          for (int j = (l[i] + d - 1) / d; j <= m / d; j++) {
            if (j >= r[i] / d)
              f[i][j] = sum[j - (l[i] + d - 1) / d] - sum[j - r[i] / d - 1];
            else
              f[i][j] = sum[j - (l[i] + d - 1) / d];
            f[i][j] = f[i][j] % mod;
          }
        }
        for (int j = 1; j <= m / d; j++) cnt = (cnt + f[n][j]) % mod;
        ans += mu[d] * cnt;
        ans = ans % mod;
        ans += mod;
        ans = ans % mod;
      }
    }
    cout << ans << endl;
    return 0;
  }
  