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 = 100002, M = 998244353;
  int n, m, t, i, j, L[N], R[N], l, r, o[N], f[N], Ans[N];
  void Add(int &x, int y) {
    x += y;
    x = x >= M ? x - M : x;
  }
  int Q(int p) {
    int i, j, k;
    t = m / p;
    for (i = 1; i <= n; i++) {
      r = R[i] / p;
      l = (L[i] + p - 1) / p;
      t -= l;
      o[i] = r - l + 1;
    }
    if (t < 0) return 0;
    for (i = 0; i <= t; i++) f[i] = 0;
    f[0] = 1;
    for (i = 1; i <= n; i++) {
      k = o[i];
      for (j = t; j >= k; j--) Add(f[j], -f[j - k] + M);
    }
    for (i = 0; i <= n; i++)
      for (j = 1; j <= t; j++) Add(f[j], f[j - 1]);
    return f[t];
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (i = 1; i <= n; i++) scanf("%d%d", L + i, R + i);
    for (i = 1; i <= m; i++) Ans[i] = Q(i);
    for (i = m; i; i--)
      for (j = i * 2; j <= m; j += i) Add(Ans[i], M - Ans[j]);
    printf("%d\n", Ans[1]);
  }
  