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>
  int n, m;
  int L[55], R[55], x[55], y[55];
  long long f[100005], g[100005];
  long long dp[100005], pre[100005];
  long long solve(int d) {
    for (int i = 1; i <= n; i++) {
      x[i] = (L[i] - 1) / d + 1;
      y[i] = R[i] / d;
    }
    for (int j = 0; j <= m / d; j++) pre[j] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < x[i]; j++) dp[j] = 0;
      for (int j = x[i]; j <= m / d; j++) {
        dp[j] = pre[j - x[i]];
        if (j > y[i]) dp[j] -= pre[j - y[i] - 1];
        dp[j] %= 998244353;
      }
      pre[0] = dp[0];
      for (int j = 1; j <= m / d; j++) pre[j] = (pre[j - 1] + dp[j]) % 998244353;
    }
    return pre[m / d];
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d%d", &L[i], &R[i]);
    for (int i = 1; i <= m; i++) f[i] = solve(i);
    for (int i = m; i >= 1; i--) {
      g[i] = f[i];
      for (int j = 2 * i; j <= m; j += i) g[i] -= g[j];
      g[i] %= 998244353;
    }
    printf("%lld\n", (g[1] + 998244353) % 998244353);
    return 0;
  }
  