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 N = 57, M = 1e5 + 7, MOD = 998244353;
  int f[N][M], s[M];
  int _L[N], _R[N], L[N], R[N];
  int primes[M], cnt, mob[M];
  bool st[M];
  void init() {
    mob[1] = 1;
    for (int i = 2; i <= M - 1; ++i) {
      if (!st[i]) primes[cnt++] = i, mob[i] = -1;
      for (int j = 0; j < cnt && 1ll * primes[j] * i < M; ++j) {
        st[primes[j] * i] = 1;
        if (i % primes[j] == 0) break;
        mob[i * primes[j]] = -mob[i];
      }
    }
  }
  int dp(int n, int m) {
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      for (int j = 0; j <= m; ++j)
        s[j] = ((j > 0 ? s[j - 1] : 0) + f[i - 1][j]) % MOD;
      for (int j = L[i]; j <= m; ++j)
        f[i][j] = (s[j - L[i]] - (j - R[i] <= 0 ? 0 : s[j - R[i] - 1])) % MOD;
    }
    int res = 0;
    for (int j = 0; j <= m; ++j) res = (res + f[n][j]) % MOD;
    return res;
  }
  int main() {
    init();
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d%d", &_L[i], &_R[i]);
    int res = 0;
    for (int d = 1; d <= m; ++d) {
      for (int i = 1; i <= n; ++i) L[i] = (_L[i] + d - 1) / d, R[i] = _R[i] / d;
      res = (res + 1ll * mob[d] * dp(n, m / d) % MOD) % MOD;
      res = (res % MOD + MOD) % MOD;
    }
    printf("%d\n", res);
    return 0;
  }
  