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 int N = 50 + 5;
  const long long M = 1e5 + 5;
  const long long mod = 998244353;
  const int inf = 1e9;
  const double eps = 1e-9;
  const double PI = acos(-1.0);
  const pair<int, int> NIL = {0, 0};
  int n, m;
  long long l[N], r[N], dp[M], sum[M], f[M];
  long long deal(long long d) {
    long long lim = m / d;
    f[0] = 1;
    for (long long i = 1; i <= lim; ++i) f[i] = 0;
    for (long long i = 1; i <= n; ++i) {
      long long L = (l[i] + d - 1) / d, R = r[i] / d;
      if (L > R) return 0;
      sum[0] = f[0];
      for (long long j = 1; j <= lim; ++j) sum[j] = (f[j] + sum[j - 1]) % mod;
      for (long long j = 0; j <= lim; ++j) {
        if (j > R)
          f[j] = (sum[j - L] - sum[j - R - 1] + mod) % mod;
        else if (j >= L)
          f[j] = sum[j - L];
        else
          f[j] = 0;
      }
    }
    long long ans = 0;
    for (long long i = 1; i <= lim; ++i) ans = (ans + f[i]) % mod;
    return ans;
  }
  void solve() {
    cin >> n >> m;
    for (long long i = 1; i <= n; ++i) cin >> l[i] >> r[i];
    for (long long d = M; d >= 1; --d) {
      dp[d] = deal(d);
      for (int i = d + d; i <= M; i += d) dp[d] = (dp[d] - dp[i] + mod) % mod;
    }
    cout << dp[1] << "\n";
  }
  int main() {
    solve();
    return 0;
  }
  