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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  long long MOD = 998244353;
  double eps = 1e-12;
  const int N = 3e5 + 7;
  int n, m;
  int mnrr[57], mxrr[57];
  void solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> mnrr[i] >> mxrr[i];
    vector<long long> ans(m + 1, 0);
    for (int g = 1; g <= m; g++) {
      long long limit = m / g;
      vector<long long> dp(limit + 1, 1);
      for (int i = 1; i <= n; i++) {
        long long rngmn = (mnrr[i] + g - 1) / g;
        long long rngmx = mxrr[i] / g;
        if (rngmn > rngmx) {
          dp[limit] = 0;
          break;
        }
        vector<long long> ndp(limit + 1, 0);
        for (int sum = rngmn; sum <= limit; sum++) {
          ndp[sum] = dp[sum - rngmn];
          if (sum > rngmx) {
            ndp[sum] = (ndp[sum] - dp[sum - rngmx - 1]) % MOD;
          }
          ndp[sum] = (ndp[sum - 1] + ndp[sum]) % MOD;
        }
        dp = ndp;
      }
      ans[g] = dp[limit];
    }
    for (int i = m; i >= 1; i--) {
      for (int j = i + i; j <= m; j += i) {
        ans[i] = (ans[i] - ans[j] + MOD) % MOD;
      }
    }
    cout << ans[1] << "\n";
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    for (int it = 1; it <= t; it++) {
      solve();
    }
    return 0;
  }
  