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 long long mod = 998244353;
  const long long N = 1e5 + 5;
  const double eps = 1e-8;
  const double PI = acos(-1);
  long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
  long long re() {
    long long x;
    scanf("%lld", &x);
    return x;
  }
  long long qk(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      a = a * a % mod;
      b /= 2;
    }
    return ans;
  }
  int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1}, dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
  int n, m;
  int l[100010], r[100010], f[100010], g[100010], a[100010];
  void solve() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d%d", l + i, r + i);
    for (int k = 1; k <= m; k++) {
      int mx = m / k;
      f[0] = 1;
      for (int i = 0; i <= mx; i++) g[i] = 1;
      for (int i = 1; i <= n; i++) {
        int mn = (l[i] - 1) / k + 1;
        for (int j = mn; j <= mx; j++) {
          f[j] = g[j - mn];
          if (j >= r[i] / k + 1)
            f[j] = ((f[j] - g[j - r[i] / k - 1]) % mod + mod) % mod;
        }
        for (int j = 0; j < mn; j++) g[j] = 0;
        for (int j = mn; j <= mx; j++) g[j] = (g[j - 1] + f[j]) % mod;
      }
      a[k] = g[mx];
    }
    for (int i = m; i; i--)
      for (int j = i * 2; j <= m; j += i)
        a[i] = ((a[i] - a[j]) % mod + mod) % mod;
    printf("%d\n", a[1]);
  }
  int main() {
    int T = 1;
    while (T--) solve();
    return 0;
  }
  