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 maxn = 1e5 + 10;
  const long long mod = 998244353;
  int l[maxn];
  int r[maxn];
  long long dp[60][maxn];
  long long cnt[maxn];
  long long sum[maxn];
  int main() {
    ios::sync_with_stdio(0);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      cin >> l[i] >> r[i];
    }
    for (int k = 1; k <= m; k++) {
      int top = m / k;
      dp[0][0] = 1;
      for (int i = 1; i <= n; i++) {
        int tl = (l[i] + k - 1) / k;
        int tr = r[i] / k;
        sum[0] = dp[i - 1][0];
        for (int j = 1; j <= top; j++) {
          sum[j] = sum[j - 1] + dp[i - 1][j];
          sum[j] %= mod;
          dp[i - 1][j] = 0;
        }
        for (int j = tl; j <= top; j++) {
          dp[i][j] += sum[j - tl] - ((j - tr - 1 < 0) ? 0 : sum[j - tr - 1]);
          dp[i][j] %= mod;
        }
      }
      for (int i = 1; i <= top; i++) {
        cnt[k] += dp[n][i];
        dp[n][i] = 0;
        cnt[k] %= mod;
      }
    }
    for (int i = m; i > 0; i--) {
      for (int j = 2; i * j <= m; j++) {
        cnt[i] -= cnt[i * j];
        cnt[i] += mod;
        cnt[i] %= mod;
      }
    }
    cout << cnt[1] << endl;
    return 0;
  }
  