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;
  unsigned long long MOD = 998244353;
  int main() {
    int N, M, gmax = -1;
    cin >> N >> M;
    vector<pair<int, int>> l(N);
    for (int i = 0; i < N; ++i) {
      cin >> l[i].first >> l[i].second;
      gmax = max(gmax, l[i].second);
    }
    vector<bool> is_prime(gmax + 1, true);
    is_prime[1] = false;
    vector<int> mu(gmax + 1, 1);
    vector<int> prime;
    for (int i = 2; i <= gmax; ++i) {
      if (is_prime[i]) {
        prime.push_back(i);
        mu[i] = -1;
      }
      for (auto p : prime) {
        if (i * p > gmax) break;
        is_prime[i * p] = false;
        if (i % p == 0) {
          mu[i * p] = 0;
          break;
        } else {
          mu[i * p] = -mu[i];
        }
      }
    }
    unsigned long long ans = 0;
    for (int i = 1; i <= gmax; ++i) {
      if (mu[i] == 0) continue;
      vector<pair<int, int>> rl(N);
      bool flag = true;
      for (int j = 0; j < N; ++j) {
        rl[j] = {(l[j].first + i - 1) / i, l[j].second / i};
        if (rl[j].first > rl[j].second) {
          flag = false;
          break;
        }
      }
      if (!flag) continue;
      int rM = M / i;
      vector<unsigned long long> ps(rM + 1, 1);
      for (int j = 0; j < N; ++j) {
        vector<unsigned long long> ps2(rM + 1, 0);
        for (int k = rl[j].first; k <= rM; ++k) {
          unsigned long long t =
              (ps[k - rl[j].first] + MOD -
               (k - rl[j].second > 0 ? ps[k - rl[j].second - 1] : 0)) %
              MOD;
          ps2[k] = (ps2[k - 1] + t) % MOD;
        }
        swap(ps, ps2);
      }
      if (mu[i] == 1) {
        (ans += ps[rM]) %= MOD;
      } else {
        (ans += MOD - ps[rM]) %= MOD;
      }
    }
    printf("%llu\n", ans);
    return 0;
  }
  