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 mod = 998244353;
  long long n, m, prim[200050], mu[200050], cnt, tt, vis[200500];
  long long dp[55][100500], l[500500], r[505000], ll[500500], rr[500500], ans,
      tot[500500];
  void prime(long long n) {
    mu[1] = 1;
    for (register int i = 2; i <= 100000; i++) {
      if (!vis[i]) prim[++cnt] = i, mu[i] = -1;
      for (int j = 1; j <= cnt && prim[j] * i <= n; j++) {
        vis[i * prim[j]] = 1;
        if (i % prim[j] == 0)
          break;
        else
          mu[i * prim[j]] = -mu[i];
      }
    }
  }
  void work(int k) {
    if (mu[k] == 0) return;
    for (int i = 1; i <= n; i++) {
      ll[i] = (l[i] - 1) / k + 1;
      rr[i] = r[i] / k;
      if (ll[i] > rr[i]) return;
    }
    int tt = m / k;
    for (int j = 0; j <= m / k; j++) tot[j] = 0;
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= m / k; j++) dp[i][j] = 0;
    }
    for (int i = ll[1]; i <= rr[1]; i++) {
      dp[1][i] = 1;
    }
    tot[0] = 0;
    for (int j = 1; j <= m / k; j++) tot[j] = tot[j - 1] + dp[1][j];
    for (int i = 2; i <= n; i++) {
      for (int j = 1; j <= tt; j++) {
        dp[i][j] =
            (tot[max(j - ll[i], 0ll)] - tot[max(j - rr[i] - 1, 0ll)] + mod) % mod;
      }
      tot[0] = 0;
      for (int j = 1; j <= m / k; j++) tot[j] = (tot[j - 1] + dp[i][j]) % mod;
    }
    for (int i = 1; i <= tt; i++) {
      ans = (ans + 1ll * mu[k] * dp[n][i] + mod) % mod;
    }
  }
  int main() {
    prime(200000);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      scanf("%lld%lld", &l[i], &r[i]);
    }
    for (int i = 1; i <= m; i++) {
      work(i);
    }
    cout << ans;
  }
  