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 int N = 20000;
  const int LG = 32 - __builtin_clz(N);
  const int INF = 1e9;
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int> > ran(n);
    for (int i = 0; i < n; i++) {
      int l, r;
      cin >> l >> r;
      ran[i] = {l, r};
    }
    int gcd = m / n;
    vector<long long> ans(gcd + 1, 0);
    for (int div = 1; div <= gcd; div++) {
      int nm = m / div;
      vector<long long> dp(nm + 1, 1);
      vector<long long> sp(nm + 2, 0);
      for (int i = n - 1; i >= 0; i--) {
        for (int j = 0; j <= nm; j++) {
          sp[j + 1] = (dp[j] + sp[j]) % MOD;
        }
        int l = (ran[i].first + div - 1) / div;
        int r = ran[i].second / div;
        if (l > r) continue;
        for (int amo = 0; amo < l; amo++) dp[amo] = 0;
        for (int amo = l; amo <= nm; amo++) {
          int top = amo - l;
          int bot = max(0, amo - r);
          dp[amo] = (sp[top + 1] - sp[bot] + MOD) % MOD;
        }
      }
      ans[div] = dp[nm];
    }
    for (int i = gcd; i > 0; i--) {
      for (int j = 2 * i; j <= gcd; j += i) {
        ans[i] = (ans[i] - ans[j] + MOD) % MOD;
      }
    }
    if (gcd == 0)
      cout << 0 << '\n';
    else
      cout << ans[1] << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    int tt = 1;
    while (tt--) solve();
    return 0;
  }
  