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;
  int n, m;
  pair<int, int> a[55];
  long long dp[55][100005], res[100005] = {0}, s[100005];
  inline void solve(int x) {
    int g = m / x;
    pair<int, int> b[55];
    for (int i = 1; i <= n; i++) {
      if (a[i].first % x == 0)
        b[i].first = a[i].first / x;
      else
        b[i].first = a[i].first / x + 1;
      b[i].second = a[i].second / x;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= g; j++) dp[i][j] = 0;
    }
    for (int i = 1; i <= g; i++) s[i] = 0;
    for (int i = b[1].first; i <= b[1].second; i++) dp[1][i] = 1;
    for (int i = 1; i <= n; i++) {
      s[0] = 0;
      for (int k = 1; k <= g; k++) {
        int r = min(b[i].second, k), l = b[i].first;
        if (r < l) continue;
        dp[i][k] = (dp[i][k] + s[k - l] - s[k - r - 1]) % (long long)(998244353);
      }
      for (int k = 1; k <= g; k++) {
        s[k] = (s[k - 1] + dp[i][k]) % (long long)(998244353);
      }
    }
    res[x] = s[g];
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      int l, r;
      cin >> l >> r;
      a[i] = {l, r};
    }
    for (int i = 1; i <= m; i++) {
      solve(i);
    }
    for (int i = m; i >= 1; i--) {
      for (int j = 2 * i; j <= m; j += i) {
        res[i] -= res[j];
        res[i] = (res[i] + (long long)(998244353)) % (long long)(998244353);
      }
    }
    cout << res[1];
  }
  