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 xm[4] = {-1, 1, 0, 0};
  int ym[4] = {0, 0, -1, 1};
  const int MOD = 998244353;
  const int MAXN = 1e5 + 5;
  const long long POW = 9973;
  int main() {
    long long n, m;
    cin >> n >> m;
    long long l[n], r[n];
    for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
    vector<long long> dp(m + 1, 0);
    for (int i = m; i >= 1; i--) {
      long long mn = 0;
      long long ar[n];
      long long ava;
      vector<vector<long long>> tdp;
      for (int j = 0; j < n; j++) {
        long long tmp;
        if (l[j] % i == 0) {
          tmp = l[j];
          mn += l[j];
        } else {
          tmp = l[j] - (l[j] % i);
          tmp += i;
          if (tmp > r[j]) goto skip;
          mn += tmp;
        }
        ar[j] = (r[j] - tmp) / i;
      }
      ava = (m - mn) / i;
      if (mn > m) goto skip;
      tdp = vector<vector<long long>>(n, vector<long long>(ava + 1, 0));
      for (int j = 0; j < n; j++) {
        if (j == 0) {
          for (int l = 0; l <= ar[j]; l++) {
            if (l > ava) break;
            tdp[0][l] = 1;
          }
          continue;
        }
        vector<long long> pref(ava + 1, 0);
        for (int l = 0; l <= ava; l++) {
          pref[l] += tdp[j - 1][l];
          pref[l] %= MOD;
          if (l + ar[j] + 1 <= ava) {
            pref[l + ar[j] + 1] =
                (pref[l + ar[j] + 1] + MOD - tdp[j - 1][l]) % MOD;
          }
        }
        long long tot = 0;
        for (int l = 0; l <= ava; l++) {
          tot += pref[l];
          tot %= MOD;
          tdp[j][l] = tot;
        }
      }
      for (int l = 0; l <= ava; l++) {
        dp[i] += tdp[n - 1][l];
        dp[i] %= MOD;
      }
      for (int j = i * 2; j <= m; j += i) dp[i] = (dp[i] + MOD - dp[j]) % MOD;
    skip:;
    }
    cout << dp[1] << "\n";
  }
  