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 maxn = 1e5 + 10;
  int l[maxn], r[maxn];
  int mu[maxn];
  int vis[maxn];
  int prim[maxn];
  int cnt = 0;
  long long mod = 998244353;
  void get_mu(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; i++) {
      if (!vis[i]) {
        prim[++cnt] = i;
        mu[i] = -1;
      }
      for (int j = 1; j <= cnt && prim[j] * i <= n; j++) {
        vis[prim[j] * i] = 1;
        if (i % prim[j] == 0)
          break;
        else
          mu[i * prim[j]] = -mu[i];
      }
    }
  }
  long long dp[55][maxn];
  long long sum[maxn];
  int main() {
    int n, m;
    get_mu(maxn - 1);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d%d", &l[i], &r[i]);
    long long ans = 0;
    for (int d = 1; d <= m; d++) {
      dp[0][0] = 1;
      long long cnt = 0;
      for (int i = 1; i <= n; i++) {
        sum[0] = dp[i - 1][0];
        for (int j = 1; j <= m / d; j++) {
          sum[j] = sum[j - 1] + dp[i - 1][j];
        }
        for (int j = ((l[i] + d - 1) / d); j <= m / d; j++) {
          if (j - r[i] / d - 1 < 0)
            dp[i][j] = sum[j - ((l[i] + d - 1) / d)];
          else
            dp[i][j] = sum[j - ((l[i] + d - 1) / d)] - sum[j - r[i] / d - 1];
          dp[i][j] %= mod;
        }
      }
      for (int j = 1; j <= m / d; j++) {
        cnt += dp[n][j];
      }
      ans += mu[d] * cnt;
      ans %= mod;
      ans += mod;
      ans %= mod;
    }
    cout << ans;
    return 0;
  }
  