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 n, m, l[60], r[60], memo[200010], L[60], R[60];
  long long dp(long long k) {
    if (memo[k] != -1) return memo[k];
    long long ans = 0;
    for (long long i = 2; i * k <= m; i++) {
      ans = (ans - dp(i * k) + mod) % mod;
    }
    long long M = m / k, esq = 0;
    for (long long i = 1; i <= n; i++) {
      L[i] = l[i] / k;
      if (l[i] % k) L[i]++;
      R[i] = r[i] / k;
      esq += L[i];
      if (R[i] < L[i]) return memo[k] = 0;
      R[i] -= L[i];
    }
    if (M < esq) return memo[k] = 0;
    M -= esq;
    vector<long long> soma(M + 10);
    for (long long i = 0; i <= min(R[1], M); i++) soma[i] = 1;
    for (long long obj = 2; obj <= n; obj++) {
      vector<long long> pref(M + 10), aux(M + 10);
      for (long long i = 0; i <= M; i++) {
        if (i != 0) pref[i] = pref[i - 1];
        aux[i] = soma[i];
        pref[i] = (pref[i] + soma[i]) % mod;
      }
      for (long long num = 0; num <= M; num++) {
        long long left = max(0ll, num - R[obj]);
        soma[num] = pref[num];
        if (left > 0) soma[num] = (soma[num] - pref[left - 1] + mod) % mod;
        ;
      }
    }
    for (long long i = 0; i <= M; i++) ans = (ans + soma[i]) % mod;
    return memo[k] = ans;
  }
  void solve() {
    memset(memo, -1, sizeof(memo));
    cin >> n >> m;
    for (long long i = 1; i <= n; i++) {
      cin >> l[i] >> r[i];
    }
    cout << dp(1) << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    solve();
  }
  