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 l[55], r[55];
  long long ans[100005];
  void solve() {
    long long n, m;
    cin >> n >> m;
    for (long long i = 1; i <= n; i++) {
      cin >> l[i] >> r[i];
    }
    for (long long x = m / n; x >= 1; x--) {
      long long sz = m / x;
      vector<long long> dp(sz + 1);
      dp[0] = 1;
      for (long long i = 1; i <= n; i++) {
        vector<long long> ndp(sz + 1);
        long long mini = 1e9, maxi = 0;
        for (long long num = 1; num * x <= r[i]; num++) {
          if (num * x < l[i]) continue;
          mini = min(mini, num);
          maxi = max(maxi, num);
        }
        for (long long j = 1; j <= sz; j++) {
          dp[j] = (dp[j] + dp[j - 1]) % mod;
        }
        for (long long j = 0; j <= sz; j++) {
          long long lower = j - maxi, upper = j - mini;
          if (upper < 0) continue;
          if (lower <= 0) {
            ndp[j] = (dp[upper]) % mod;
            continue;
          }
          ndp[j] = (dp[upper] - dp[lower - 1] + mod) % mod;
        }
        dp = ndp;
      }
      for (auto u : dp) ans[x] = (ans[x] + u);
      for (long long num = 2 * x; num <= m; num += x) {
        ans[x] = (ans[x] - ans[num] + mod) % mod;
      }
    }
    cout << ans[1];
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    long long t = 1;
    while (t--) {
      solve();
    }
  }
  