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 INF = 2e9;
  inline int Max(const int &a, const int &b) { return a > b ? a : b; }
  inline int Min(const int &a, const int &b) { return a > b ? b : a; }
  inline int read() {
    int f = 1, x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    return x * f;
  }
  const int N = 55;
  const int M = 1e5 + 7;
  const int mod = 998244353;
  int n, m;
  int l[N], r[N];
  int mu[M];
  int prime[M], ntp[M], cnt;
  void xxs(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; ++i) {
      if (!ntp[i]) {
        prime[++cnt] = i;
        mu[i] = -1;
      }
      for (int j = 1; j <= cnt; ++j) {
        if (i * prime[j] > n) break;
        ntp[i * prime[j]] = 1;
        if (i % prime[j] == 0) {
          mu[i * prime[j]] = 0;
          break;
        }
        mu[i * prime[j]] = -mu[i];
      }
    }
  }
  long long dp[M], s[M], ans;
  int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; ++i) l[i] = read(), r[i] = read();
    xxs(m);
    for (int i = 1; i <= m; ++i) {
      if (mu[i] == 0) continue;
      int lim = m / i;
      for (int j = 1; j <= lim; ++j) dp[j] = 0;
      dp[0] = 1;
      for (int j = 1; j <= n; ++j) {
        int L = (l[j] + i - 1) / i, R = r[j] / i;
        for (int k = 0; k <= lim; ++k) {
          s[k] = (k > 0 ? s[k - 1] : 0);
          s[k] = (s[k] + dp[k]) % mod;
        }
        for (int k = 0; k <= lim; ++k) {
          dp[k] = (k - L >= 0 ? s[k - L] : 0);
          dp[k] = (dp[k] - (k - R - 1 >= 0 ? s[k - R - 1] : 0) + mod) % mod;
        }
      }
      long long res = 0;
      for (int k = 0; k <= lim; ++k) res = (res + dp[k]) % mod;
      ans = (ans + res * mu[i] + mod) % mod;
    }
    printf("%lld\n", ans);
    return 0;
  }
  