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;
  using ll = long long;
  using ld = long double;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  const char nl = '\n';
  const ll mod = 998244353;
  const int max_n = 50;
  const int max_m = 1e5;
  const int max_sum = max_n * max_m;
  int l[max_n + 1], r[max_n + 1];
  ll f[max_m + 1];
  ll cnt[max_m + 2][max_n + 1];
  ll cnt_pref[max_m + 2][max_n + 1];
  void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
    int l_sum = 0;
    for (int i = 1; i <= n; i++) l_sum += l[i];
    if (l_sum > m) {
      cout << 0 << nl;
      return;
    }
    for (int d = m / n; d >= 1; d--) {
      int ms = m / d;
      bool can = true;
      for (int i = 1; i <= n; i++) {
        can &= (l[i] + d - 1) / d * d <= r[i];
      }
      if (!can) continue;
      for (int i = 0; i <= n; i++) {
        for (int s = 0; s <= ms; s++) {
          cnt[s][i] = cnt_pref[s][i] = 0;
        }
      }
      cnt[0][0] = 1;
      for (int s = 0; s <= ms; s++) cnt_pref[s][0] = 1;
      for (int i = 1; i <= n; i++) {
        for (int s = 0; s <= ms; s++) {
          int left = max(0, s - r[i] / d), right = s - (l[i] + d - 1) / d;
          if (left > right) continue;
          ll range_sum = cnt_pref[right][i - 1];
          if (left > 0) {
            range_sum -= cnt_pref[left - 1][i - 1];
            range_sum += mod;
            range_sum %= mod;
          }
          cnt[s][i] = range_sum;
          cnt_pref[s][i] = cnt[s][i];
          if (s > 0) {
            cnt_pref[s][i] += cnt_pref[s - 1][i];
            cnt_pref[s][i] %= mod;
          }
        }
      }
      f[d] = cnt_pref[ms][n];
      for (int d2 = 2 * d; d2 <= m; d2 += d) {
        f[d] -= f[d2];
        f[d] += mod;
        f[d] %= mod;
      }
    }
    cout << f[1] << nl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    while (t--) solve();
    return 0;
  }
  