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 int N = 1e5 + 7, M = 50 + 7, mod = 998244353;
  int prime[N], mu[N], l[M], r[M], r_[M], sum_[N], sum[N], dp[M][N];
  bool p[N];
  inline void init(int n) {
    int cnt = 0;
    p[0] = p[1] = true;
    mu[1] = 1;
    for (register int i = 2; i <= n; i++) {
      if (!p[i]) {
        prime[++cnt] = i;
        mu[i] = -1;
      }
      for (register int j = 1; j <= cnt && i * prime[j] <= n; j++) {
        int t = i * prime[j];
        p[t] = true;
        if (i % prime[j] == 0) {
          mu[t] = 0;
          break;
        }
        mu[t] = -mu[i];
      }
    }
  }
  inline int min(int a, int b) { return a < b ? a : b; }
  inline int do_dp(int n, int m) {
    int ans = 0;
    for (register int i = 0; i <= m; i++) {
      sum_[i] = 1;
    }
    for (register int i = 1; i <= n; i++) {
      for (register int j = 0; j <= m; j++) {
        int t = min(r_[i], j);
        if (i == n) ans = (ans + sum_[t]) % mod;
        sum[j] = sum_[t];
        if (j > 0) sum[j] = (sum[j] + sum[j - 1]) % mod;
      }
      for (register int j = 0; j <= m; j++) {
        sum_[j] = sum[j];
      }
    }
    return ans;
  }
  int main() {
    int n, m, mi, ans = 0;
    scanf("%d %d", &n, &m);
    mi = m + 1;
    init(m);
    for (register int i = 1; i <= n; i++) {
      scanf("%d %d", &l[i], &r[i]);
    }
    for (register int i = 1; i <= m; i++) {
      int sum = 0, tm = m / i;
      for (register int k = 1; k <= n; k++) {
        int tl = (l[k] - 1) / i + 1;
        r_[k] = r[k] / i - tl;
        sum = min(sum + tl, mi);
      }
      if (tm >= sum) ans = ((ans + mu[i] * do_dp(n, tm - sum)) % mod + mod) % mod;
    }
    printf("%d", ans);
    return 0;
  }
  