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 maxn = 1e5 + 10, mod = 998244353;
  int mu[maxn], n, m, l[51], r[51], prime[maxn], cnt;
  bool is_composite[maxn];
  int dp[maxn], pre[maxn];
  int cal(int d) {
    int M = m / d;
    for (int i = 1; i <= M; i++) dp[i] = 0;
    dp[0] = 1;
    for (int i = 1; i <= n; i++) {
      int L = (l[i] + d - 1) / d, R = r[i] / d;
      if (L > R) return 0;
      pre[0] = dp[0];
      for (int j = 1; j <= M; j++) pre[j] = (pre[j - 1] + dp[j]) % mod;
      for (int j = 0; j <= M; j++) {
        dp[j] = (((j - L >= 0) ? pre[j - L] : 0) -
                 ((j - R - 1 >= 0) ? pre[j - R - 1] : 0)) %
                mod;
      }
    }
    pre[0] = dp[0];
    for (int j = 0; j <= M; j++) pre[j] = (pre[j - 1] + dp[j]) % mod;
    return pre[M];
  }
  int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) scanf("%d%d", &l[i], &r[i]);
    mu[1] = 1;
    for (int i = 2; i <= m; i++) {
      if (!is_composite[i]) {
        prime[++cnt] = i;
        mu[i] = mod - 1;
      }
      for (int j = 1; j <= cnt && i * prime[j] <= m; j++) {
        is_composite[i * prime[j]] = true;
        if (i % prime[j])
          mu[i * prime[j]] = (mod - mu[i]) % mod;
        else
          break;
      }
    }
    int ans = 0;
    for (int d = 1; d <= m; d++) {
      ans = (ans + 1LL * cal(d) * mu[d]) % mod;
    }
    cout << ans << '\n';
  }
  