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;
  inline long long read() {
    long long x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9')
      x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
    return x * f;
  }
  const long long mod = 998244353, N = 1e5 + 5;
  long long p[N], chk[N], tot, mu[N];
  void init(long long n) {
    mu[1] = 1;
    for (long long i = 2; i <= n; i++) {
      if (!chk[i]) p[++tot] = i, mu[i] = -1;
      for (long long j = 1; j <= tot && p[j] * i <= n; j++) {
        chk[i * p[j]] = 1;
        if (!(i % p[j])) break;
        mu[i * p[j]] = -mu[i];
      }
    }
  }
  long long ans;
  long long l[N], r[N], f[N], s[N];
  long long n, m;
  signed main() {
    init(N - 5);
    n = read(), m = read();
    for (long long i = 1; i <= n; i++) l[i] = read(), r[i] = read();
    for (long long d = 1; d <= m; d++) {
      long long lim = m / d;
      for (long long i = 1; i <= lim; i++) f[i] = 0;
      f[0] = s[0] = 1;
      for (long long j = 1; j <= lim; j++) s[j] = (s[j - 1] + f[j]) % mod;
      for (long long i = 1; i <= n; i++) {
        long long L = (l[i] - 1) / d + 1, R = r[i] / d;
        for (long long j = 1; j <= lim; j++)
          f[j] = ((j - L < 0ll ? 0ll : s[max(j - L, 0ll)]) -
                  (j - R - 1 < 0ll ? 0ll : s[max(j - R - 1, 0ll)])) %
                 mod;
        f[0] = s[0] = 0;
        for (long long j = 1; j <= lim; j++) s[j] = (s[j - 1] + f[j]) % mod;
      }
      ans += mu[d] * s[lim];
      ans %= mod;
    }
    printf("%lld\n", (ans + mod) % mod);
    return 0;
  }
  