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 long long mod = 998244353;
  long long calc(vector<pair<long long, long long>> a, long long sum) {
    long long n = a.size();
    long long dp[n + 1][sum + 1];
    for (long long i = 0; i < n + 1; ++i) {
      for (long long j = 0; j < sum + 1; ++j) {
        dp[i][j] = 0;
      }
    }
    long long pref[n + 1][sum + 2];
    for (long long k = 0; k < n + 1; ++k) {
      for (long long i = 0; i < sum + 2; ++i) {
        pref[k][i] = 0;
      }
    }
    dp[0][0] = 1;
    pref[0][0] = 0;
    for (long long l = 1; l < sum + 2; ++l) {
      pref[0][l] = (pref[0][l - 1] + dp[0][l - 1]) % mod;
    }
    for (long long i = 1; i <= n; ++i) {
      for (long long s = 0; s <= sum; ++s) {
        dp[i][s] += pref[i - 1][max(0ll, s - a[i - 1].first + 1)] -
                    pref[i - 1][max(0ll, s - a[i - 1].second)];
        dp[i][s] %= mod;
      }
      pref[i][0] = 0;
      for (long long l = 1; l < sum + 2; ++l) {
        pref[i][l] = pref[i][l - 1] + dp[i][l - 1];
        pref[i][l] %= mod;
      }
    }
    long long ans = 0;
    for (long long m = 0; m <= sum; ++m) {
      ans += dp[n][m];
      ans %= mod;
    }
    return ans;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    long long n, m;
    cin >> n >> m;
    vector<pair<long long, long long>> a;
    for (long long i = 0; i < n; ++i) {
      long long l, r;
      cin >> l >> r;
      a.push_back({l, r});
    }
    long long ans[m + 1];
    for (long long g = 1; g <= m; ++g) {
      vector<pair<long long, long long>> b;
      for (long long i = 0; i < n; ++i) {
        b.push_back({(a[i].first + g - 1) / g, a[i].second / g});
      }
      ans[g] = calc(b, m / g);
    }
    long long real_ans[m + 1];
    for (long long j = m; j >= 1; --j) {
      real_ans[j] = ans[j];
      for (long long i = j * 2; i <= m; i += j) {
        real_ans[j] -= real_ans[i];
        real_ans[j] %= mod;
      }
    }
    cout << (real_ans[1] + mod) % mod;
  }
  