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>
  const long long N = 55, M = 100005, P = 998244353;
  long long n, m, l[N], r[N], f[N][M], ans[M];
  signed main() {
    std::cin >> n >> m;
    for (long long i = 1; i <= n; i++) std::cin >> l[i] >> r[i];
    for (long long g = m / n; g >= 1; g--) {
      for (long long i = 0; i <= n; i++)
        for (long long j = 0; j <= m / g + 1; j++) f[i][j] = 0;
      for (long long j = 0; j <= m / g + 1; j++) f[0][j] = j + 1;
      for (long long i = 1; i <= n; i++) {
        long long L = (l[i] + g - 1) / g, R = r[i] / g;
        for (long long a = 0; a <= m / g + 1; a++) {
          long long cl = a + L, cr = std::min(a + R, m / g);
          if (cl > cr)
            f[i][a] = 0;
          else
            f[i][a] = (f[i - 1][cr] - (cl ? f[i - 1][cl - 1] : 0) + P) % P;
          if (a > 0) f[i][a] = (f[i][a] + f[i][a - 1]) % P;
        }
      }
      ans[g] = f[n][0];
      for (long long i = 2 * g; i <= m; i += g)
        ans[g] = (ans[g] - ans[i] + P) % P;
    }
    std::cout << ans[1];
  }
  