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 int N = 1e5;
  const int M = 50;
  int num[N + 5];
  int mob[N + 5];
  void pre() {
    mob[1] = 1;
    for (int i = 2; i <= N; i++) {
      if (num[i] == 0) {
        for (int j = i; j <= N; j += i) {
          mob[j]++;
          num[j] = j;
        }
        long long x = i;
        x = x * i;
        for (long long j = x; j <= N; j += x) mob[j] = -N;
      }
      if (mob[i] < 0)
        mob[i] = 0;
      else if (mob[i] % 2)
        mob[i] = -1;
      else
        mob[i] = 1;
    }
  }
  long long dp[M + 1][N + 1];
  long long ssum[N + 5];
  int l[M + 1], r[M + 1];
  int nr[M + 1];
  inline long long get(int l, int r) {
    if (l < 0) l = 0;
    return (ssum[l] - ssum[r + 1] + MOD) % MOD;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    pre();
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
    long long res = 0;
    for (int i = 1; i <= m; i++) {
      int sisa = m / i;
      bool bisa = true;
      for (int j = 1; j <= n; j++) {
        int nl = l[j] / i;
        if (nl * i < l[j]) nl++;
        nr[j] = r[j] / i;
        if (nr[j] * i < l[j] || nl * i > r[j] || nl > nr[j]) bisa = false;
        sisa -= nl;
        nr[j] -= nl;
      }
      if (mob[i] == 0 || !bisa || sisa < 0) continue;
      for (int j = 0; j <= max(sisa, nr[j]) + 1; j++) ssum[j] = 0;
      for (int j = min(sisa, nr[1]); j >= 0; j--) {
        dp[1][j] = 1;
        ssum[j] = (ssum[j + 1] + dp[1][j]) % MOD;
      }
      for (int j = 2; j <= n; j++) {
        for (int k = sisa; k >= 0; k--) {
          dp[j][k] = get(k - nr[j], k);
        }
        for (int k = sisa; k >= 0; k--) {
          ssum[k] = (ssum[k + 1] + dp[j][k]) % MOD;
        }
      }
      res = (res + MOD + mob[i] * ssum[0]) % MOD;
    }
    cout << res << '\n';
  }
  