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 N = 1e5 + 7, mod = 998244353;
  int n, m, mx, mn, l[N], r[N], f[N], g[N], a[N], i, j, k;
  int main() {
    for (scanf("%d%d", &n, &m), i = 1; i <= n; ++i) scanf("%d%d", l + i, r + i);
    for (k = 1; k <= m; ++k) {
      for (mx = m / k, f[0] = 1, i = 0; i <= mx; ++i) g[i] = 1;
      for (i = 1; i <= n; ++i) {
        for (j = mn = (l[i] - 1) / k + 1; j <= mx; ++j) {
          f[j] = g[j - mn];
          if (j >= r[i] / k + 1)
            f[j] = ((f[j] - g[j - r[i] / k - 1]) % mod + mod) % mod;
        }
        for (j = 0; j < mn; ++j) g[j] = 0;
        for (j = mn; j <= mx; ++j) g[j] = (g[j - 1] + f[j]) % mod;
      }
      a[k] = g[mx];
    }
    for (i = m; i; --i)
      for (j = i * 2; j <= m; j += i) a[i] = ((a[i] - a[j]) % mod + mod) % mod;
    return 0 * printf("%d\n", a[1]);
  }
  