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 = 1e5 + 5;
  const int mod = 998244353;
  inline int add(int x, int y) { return (x + y >= mod ? x + y - mod : x + y); }
  inline int sub(int x, int y) { return (x - y < 0 ? x - y + mod : x - y); }
  inline int gun(int x, int y) { return ((x * 1ll * y) % mod); }
  int n, m;
  pair<int, int> a[55];
  int g[N];
  int dp[N][55];
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
      cin >> a[i].first >> a[i].second;
    }
    for (int now = 1; now <= m; now++) {
      int sum = m / now;
      for (int idx = n; idx >= 0; idx--) {
        int l = (a[idx].first + now - 1) / now;
        int r = a[idx].second / now;
        for (int cur = 0; cur <= sum; cur++) {
          if (idx == n) {
            dp[cur][idx] = 1;
            continue;
          }
          int ans = 0;
          if (l <= r && cur + l <= sum) {
            int ses = min(cur + r, sum);
            ans = dp[ses][idx + 1];
            if (cur + l - 1 >= 0) ans = sub(ans, dp[cur + l - 1][idx + 1]);
          }
          dp[cur][idx] = ans;
        }
        for (int cur = 1; cur <= sum; cur++) {
          dp[cur][idx] = add(dp[cur][idx], dp[cur - 1][idx]);
        }
      }
      g[now] = dp[0][0];
      ;
    }
    for (int i = m; i >= 1; i--) {
      for (int j = i + i; j <= m; j += i) {
        g[i] = sub(g[i], g[j]);
      }
    }
    cout << g[1] << "\n";
    return 0;
  }
  