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;
  void solve() {
    long long n, m;
    cin >> n >> m;
    vector<long long> l(n), r(n);
    for (long long i = 0; i < n; i++) cin >> l[i] >> r[i];
    vector<long long> result(m + 1);
    for (long long g = 1; g <= m; g++) {
      long long to = m / g;
      vector<long long> dp(to + 1, 1);
      for (long long i = 0; i < n; i++) {
        long long mn = (l[i] + g - 1) / g;
        long long mx = r[i] / g;
        if (mx < mn) {
          dp[to] = 0;
          break;
        }
        vector<long long> ndp(to + 1);
        for (long long s = mn; s <= to; s++) {
          ndp[s] =
              (ndp[s - 1] + dp[s - mn] - (s - mx - 1 >= 0 ? dp[s - mx - 1] : 0)) %
              MOD;
          if (ndp[s] < 0) ndp[s] += MOD;
        }
        dp = ndp;
      }
      result[g] = dp[to];
    }
    for (long long g = m; g; g--) {
      for (long long i = 2 * g; i <= m; i += g) {
        result[g] -= result[i];
        if (result[g] < 0) result[g] += MOD;
      }
    }
    cout << result[1];
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    solve();
    return 0;
  }
  