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;
  const int maxn = 1e5 + 5;
  const int mod = 998244353;
  int cnt, prime[maxn], mu[maxn], n, m;
  int l[maxn], r[maxn];
  ll dp[maxn], sum[maxn], ans;
  bitset<maxn> vis;
  ll solve(int d) {
    int M = m / d, L, R;
    for (int i = 1; i <= M; i++) dp[i] = 0;
    dp[0] = 1;
    for (int i = 1; i <= n; i++) {
      L = (l[i] + d - 1) / d, R = r[i] / d;
      if (L > R) return 0;
      sum[0] = dp[0];
      for (int j = 1; j <= M; j++) sum[j] = sum[j - 1] + dp[j];
      for (int j = 0; j <= M; j++)
        dp[j] = ((j - L >= 0 ? sum[j - L] : 0) + mod -
                 (j - R - 1 >= 0 ? sum[j - R - 1] : 0)) %
                mod;
    }
    ll ret = 0;
    for (int i = 1; i <= M; i++) ret = (ret + dp[i]) % mod;
    return ((ret % mod) + mod) % mod;
  }
  int main() {
    int N = 1e5;
    mu[1] = 1;
    for (int i = 2; i <= N; i++) {
      if (!vis[i]) prime[++cnt] = i, mu[i] = -1;
      for (int j = 1; j <= cnt && i * prime[j] <= N; j++) {
        vis[i * prime[j]] = 1;
        if (i % prime[j] == 0) break;
        mu[i * prime[j]] = -mu[i];
      }
    }
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d %d", &l[i], &r[i]);
    for (int i = 1; i <= m; i++) ans = (ans + 1ll * solve(i) * mu[i]) % mod;
    printf("%lld\n", ((ans % mod) + mod) % mod);
    return 0;
  }
  