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>
  const long long INF = 0x3f3f3f3f;
  const long long M = 1e5 + 5;
  const long long mod = 998244353;
  long long n, m, cnt, ans;
  long long vis[M], prime[M], mu[M];
  long long lc[M], rc[M], l[M], r[M], f[M], sum[M];
  inline void Init() {
    mu[1] = 1;
    for (long long i = 2; i < M; ++i) {
      if (!vis[i]) prime[++cnt] = i, mu[i] = -1;
      for (long long j = 1; j <= cnt && prime[j] * i < M; ++j) {
        vis[prime[j] * i] = 1;
        if (i % prime[j] == 0) {
          mu[prime[j] * i] = 0;
          break;
        }
        mu[prime[j] * i] = -mu[i];
      }
    }
  }
  inline long long Solve(long long m) {
    for (long long i = 0; i <= m; ++i) f[i] = sum[i] = 0;
    for (long long i = 1; i <= n; ++i) {
      if (i == 1) {
        for (long long j = lc[i]; j <= rc[i]; ++j) f[j] = 1;
        for (long long j = 1; j <= m; ++j) sum[j] = sum[j - 1] + f[j];
      } else {
        for (long long j = 1; j <= m; ++j)
          f[j] = (sum[std ::max(j - lc[i], 0ll)] -
                  sum[std ::max(j - rc[i] - 1, 0ll)] + mod) %
                 mod;
        for (long long j = 1; j <= m; ++j) sum[j] = (sum[j - 1] + f[j]) % mod;
      }
    }
    long long ans = 0;
    for (long long i = 0; i <= m; ++i) ans = (ans + f[i]) % mod;
    return ans;
  }
  inline long long read() {
    long long f = 1, s = 0;
    char ch = getchar();
    while (!isdigit(ch)) (ch == '-') && (f = -1), ch = getchar();
    while (isdigit(ch)) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return f * s;
  }
  signed main() {
    n = read(), m = read();
    Init();
    for (long long i = 1; i <= n; ++i) l[i] = read(), r[i] = read();
    for (long long d = 1; d <= m; ++d) {
      if (!mu[d]) continue;
      long long mark = 0;
      for (long long j = 1; j <= n; ++j)
        lc[j] = std ::max((l[j] + d - 1) / d, 1ll), rc[j] = r[j] / d;
      for (long long j = 1; j <= n; ++j)
        if (lc[j] > rc[j]) {
          mark = 1;
          break;
        }
      if (mark) continue;
      ans = (ans + mu[d] * Solve(m / d) % mod + mod) % mod;
    }
    std ::cout << ans;
    return 0;
  }
  