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 N = 1e5 + 5;
  int n, m;
  int f[51][N];
  unsigned long long sum[N];
  int l[N], r[N];
  unsigned long long g[N];
  void add(unsigned long long &a, unsigned long long b) {
    a += b;
    if (a >= mod) a -= mod;
  }
  int main() {
    cin >> n >> m;
    for (int i = (1); i < (n + 1); i++) scanf("%d%d", &l[i], &r[i]);
    for (int gc = (1); gc < (m + 1); gc++) {
      int tot = m / gc;
      for (int i = (1); i < (tot + 1); i++) f[1][i] = 0, sum[i] = 0;
      for (int i = ((l[1] + gc - 1) / gc); i < (r[1] / gc + 1); i++) f[1][i] = 1;
      for (int i = (1); i < (tot + 1); i++) sum[i] = (sum[i - 1] + f[1][i]) % mod;
      for (int i = (2); i < (n + 1); i++) {
        int mi = (l[i] + gc - 1) / gc, mx = r[i] / gc;
        if (mi > mx) {
          sum[tot] = 0;
          break;
        }
        for (int j = (mi); j < (tot + 1); j++) {
          if (j <= mx)
            f[i][j] = sum[j - mi];
          else
            f[i][j] = (sum[j - mi] - sum[j - mx - 1] + mod) % mod;
        }
        sum[0] = 0;
        for (int j = (1); j < (tot + 1); j++)
          sum[j] = (sum[j - 1] + f[i][j]) % mod;
      }
      g[gc] = sum[tot];
    }
    for (int i = (m + 1) - 1; i >= (1); i--) {
      for (int j = i + i; j <= m; j += i) g[i] = (g[i] - g[j] + mod) % mod;
    }
    printf("%lld\n", g[1]);
    return 0;
  }
  