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;
  int i, j, k, n, m, t, l[55], r[55], cnt;
  int g[100500], f[100500][55], pre[100500][55];
  int su(int a, int b) {
    a += b;
    return (a >= 998244353) ? a - 998244353 : a;
  }
  void get(int x) {
    int i, j, k, res = 0, l1, r1;
    for (i = x; i <= m; i += x) {
      memset(f[i], 0, sizeof(f[i]));
      memset(pre[i], 0, sizeof(pre[i]));
    }
    for (i = 1; i <= n; i++) {
      r1 = (r[i] / x) * x;
      l1 = ((l[i] + x - 1) / x) * x;
      if (l1 > r1) {
        return;
      }
      if (i == 1) {
        for (j = l1; j <= r1; j += x) {
          f[j][i] = 1;
        }
      } else {
        for (j = x; j <= m; j += x) {
          cnt++;
          if (j <= l1) {
            continue;
          }
          if (j <= r1 + x) {
            f[j][i] = pre[j - l1][i - 1];
          } else {
            f[j][i] = su(pre[j - l1][i - 1], 998244353 - pre[j - r1 - x][i - 1]);
          }
        }
      }
      for (j = x; j <= m; j += x) {
        pre[j][i] = su(pre[j - x][i], f[j][i]);
      }
    }
    for (i = x; i <= m; i += x) {
      res = su(res, f[i][n]);
    }
    g[x] = res;
  }
  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++) {
      get(i);
    }
    for (i = m; i >= 1; i--) {
      for (j = i + i; j <= m; j += i) {
        g[i] = su(g[i], 998244353 - g[j]);
      }
    }
    printf("%d", g[1]);
  }
  