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 long long N = 1e5 + 1e3 + 7, P = 998244353;
  long long n, m;
  long long f[51][N], g[N], l[51], r[51];
  long long get(long long i, long long j) { return j < 0 ? 0 : f[i][j]; }
  signed main() {
    scanf("%lld%lld", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%lld%lld", &l[i], &r[i]);
    for (long long d = m; d >= 1; d--) {
      long long t = m / d;
      for (long long j = 0; j <= t; j++) f[0][j] = 1;
      for (long long j = 1; j <= n; j++) {
        long long L = (l[j] + d - 1) / d, R = r[j] / d;
        for (long long k = 0; k <= t; k++)
          f[j][k] =
              (f[j][k - 1] + get(j - 1, k - L) - get(j - 1, k - R - 1) + P) % P;
      }
      g[d] = f[n][t];
      for (long long j = 2; j <= t; j++) g[d] = (g[d] - g[j * d] + P) % P;
    }
    printf("%lld\n", g[1]);
  }
  