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);
  const long long MAX_VALEURS = (200 * 1000);
  const long long MAX_ETOILES = (50);
  long long Ans[MAX_VALEURS];
  long long DP[MAX_VALEURS];
  pair<long long, long long> Bornes[MAX_ETOILES];
  long long nbEtoiles, borneSup;
  long long FindDP(long long d) {
    long long borne = borneSup / d;
    fill_n(DP + 1, borne, 0LL);
    DP[0] = 1;
    for (long long i = 0; i < nbEtoiles; i++) {
      long long sum = 0;
      long long inf = (Bornes[i].first + d - 1) / d, sup = Bornes[i].second / d;
      for (long long j = inf; j <= sup; j++) {
        sum += DP[borne - j];
      }
      sum %= MOD;
      for (long long j = borne; j >= 0; j--) {
        DP[j] = sum;
        if (j >= inf) sum -= DP[j - inf];
        if (j > sup) sum += DP[j - sup - 1];
        sum %= MOD;
        if (sum < 0) sum += MOD;
      }
    }
    long long ans = 0;
    for (long long i = 0; i <= borne; i++) {
      ans += DP[i];
    }
    return ans % MOD;
  }
  void Solve() {
    scanf("%lld %lld", &nbEtoiles, &borneSup);
    for (long long i = 0; i < nbEtoiles; i++) {
      scanf("%lld %lld", &Bornes[i].first, &Bornes[i].second);
    }
    for (long long i = 1; i <= borneSup; i++) {
      Ans[i] = FindDP(i);
    }
    for (long long i = borneSup; i > 0; i--) {
      for (long long j = 2 * i; j <= borneSup; j += i) {
        Ans[i] -= Ans[j];
      }
      Ans[i] %= MOD;
      if (Ans[i] < 0) Ans[i] += MOD;
    }
    printf("%lld\n", Ans[1]);
    return;
  }
  signed main() {
    long long nbTests = 1;
    while (nbTests--) {
      Solve();
    }
    return 0;
  }
  