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 mod = 998244353;
  int main() {
    int n, m;
    cin >> n >> m;
    vector<int> l(n);
    vector<int> r(n);
    for (int i = 0; i < n; ++i) {
      cin >> l[i] >> r[i];
    }
    vector<long long> ans(m + 1);
    for (int d = 1; d <= m; ++d) {
      vector<vector<long long>> dp(n + 1, vector<long long>(m / d + 1));
      vector<long long> pre(m / d + 1);
      dp[0][0] = 1;
      for (int i = 0; i < n; ++i) {
        int nl = (l[i] + d - 1) / d;
        int nr = r[i] / d;
        assert(nl > 0);
        if (nl > nr) {
          continue;
        }
        pre[0] = dp[i][0];
        for (int j = 1; j <= m / d; ++j) {
          pre[j] = (pre[j - 1] + dp[i][j]) % mod;
        }
        for (int j = nl; j <= m / d; ++j) {
          dp[i + 1][j] += pre[j - nl];
          dp[i + 1][j] += mod - (j - nr - 1 >= 0 ? pre[j - nr - 1] : 0);
          dp[i + 1][j] %= mod;
        }
      }
      ans[d] = accumulate(dp[n].begin(), dp[n].end(), 0ll) % mod;
    }
    for (int i = m; i >= 1; --i) {
      for (int j = i + i; j <= m; j += i) {
        ans[i] += mod - ans[j];
        ans[i] %= mod;
      }
    }
    cout << ans[1] << '\n';
  }
  