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 oo = 1e18;
  const long long mod = 998244353;
  int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    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];
    }
    auto modfix = [&](long long &x) {
      if (x >= mod)
        x -= mod;
      else if (x < 0)
        x += mod;
    };
    vector<long long> res(m + 1, 0);
    for (long long g = 1; g <= m; g++) {
      long long mm = m / g;
      vector<vector<long long>> pd(n + 1, vector<long long>(mm + 1, 1));
      for (long long i = n - 1; i >= 0; i--) {
        for (long long second = 1; second <= mm; second++) {
          pd[i + 1][second] += pd[i + 1][second - 1];
          modfix(pd[i + 1][second]);
        }
        for (long long second = 0; second <= mm; second++) {
          pd[i][second] = 0;
          long long a = second + ceil(1.0 * l[i] / g), b = second + r[i] / g;
          b = min(b, mm);
          if (a > b) continue;
          pd[i][second] = pd[i + 1][b] - pd[i + 1][a - 1];
          modfix(pd[i][second]);
        }
      }
      res[g] = pd[0][0];
    }
    for (long long g = m; g >= 1; g--) {
      for (long long mult = g + g; mult <= m; mult += g) {
        res[g] -= res[mult];
        modfix(res[g]);
      }
    }
    cout << res[1] << endl;
    return 0;
  }
  