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 maxn = 1e5 + 5;
  int cnt = 0, vis[maxn], prime[maxn], mu[maxn];
  int f[55][maxn], l[55], r[55], l1[55], r1[55];
  void init(int n) {
    vis[1] = mu[1] = 1;
    for (int i = (2); i <= (n); i++) {
      if (!vis[i]) prime[++cnt] = i, mu[i] = -1;
      for (int j = (1); j <= (cnt); j++) {
        if (i * prime[j] > n) break;
        vis[i * prime[j]] = 1;
        if (i % prime[j] == 0) {
          mu[i * prime[j]] = 0;
          break;
        }
        mu[i * prime[j]] = -mu[i];
      }
    }
  }
  int n, m;
  int calc(int x) {
    long long sum = 0;
    for (int i = (1); i <= (n); i++) {
      l[i] = (l1[i] - 1) / x + 1;
      r[i] = r1[i] / x;
      sum += l[i];
      if (i == 1)
        for (int j = (l[i]); j <= (r[i]); j++) f[1][j] = 1;
      else {
        for (int j = (sum); j <= (m / x); j++) {
          int L = max(1, j - r[i]), R = j - l[i];
          f[i][j] = ((f[i - 1][R] - f[i - 1][L - 1]) % mod + mod) % mod;
        }
      }
      for (int j = (1); j <= (m / x); j++) (f[i][j] += f[i][j - 1]) %= mod;
    }
    return f[n][m / x];
  }
  int main() {
    init(1e5);
    scanf("%d", &n);
    scanf("%d", &m);
    int mnr = 1e9;
    for (int i = (1); i <= (n); i++) {
      scanf("%d", &l1[i]), scanf("%d", &r1[i]);
      mnr = min(mnr, r1[i]);
    }
    long long ans = 0;
    for (int x = (1); x <= (mnr); x++) {
      ans = ((ans + calc(x) * mu[x]) % mod + mod) % mod;
    }
    cout << ans;
  }
  