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 mod = 998244353;
  const int N = 2e5 + 5;
  long long prime[N + 5], mu[N + 5], tot = 0;
  bool vis[N + 5];
  long long n, m;
  long long l[55], r[55];
  long long f[N], g[N];
  long long ans = 0;
  void init() {
    mu[1] = 1;
    for (long long i = 2; i <= N; i++) {
      if (!vis[i]) {
        prime[tot] = i;
        tot++;
        mu[i] = -1;
      }
      for (long long j = 0; j < tot && prime[j] * i <= N; j++) {
        vis[prime[j] * i] = 1;
        if (i % prime[j] == 0) {
          mu[prime[j] * i] = 0;
          break;
        }
        mu[prime[j] * i] = -mu[i];
      }
    }
  }
  int main() {
    init();
    scanf("%lld%lld", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%lld%lld", l + i, r + i);
    for (long long i = 1; i <= m; i++) {
      f[0] = g[0] = 1;
      for (long long j = 1; j <= m / i; j++) f[j] = 0, g[j] = 1;
      for (long long j = 1; j <= n; j++) {
        long long ql = (l[j] - 1) / i + 1, qr = r[j] / i;
        if (ql > qr) {
          g[m / i] = 0;
          break;
        }
        for (int k = 0; k < ql; k++) f[k] = 0;
        for (int k = ql; k <= m / i; k++)
          f[k] = g[k - ql] - (k >= qr + 1 ? g[k - qr - 1] : 0);
        g[0] = f[0];
        for (int k = 1; k <= m / i; k++) g[k] = (g[k - 1] + f[k]) % mod;
      }
      ans += mu[i] * g[m / i] % mod;
      ans %= mod;
    }
    printf("%lld\n", (ans % mod + mod) % mod);
    return 0;
  }
  