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;
  long long n, A[71], B[71], L[71], R[71], C[71], mm, s, DPF[100071],
      DP[71][100071];
  void solve(long long t) {
    long long m = mm / t;
    s = 0;
    for (long long i = 1; i <= n; ++i) {
      L[i] = (A[i] - 1) / t + 1;
      R[i] = B[i] / t;
      s += L[i];
      C[i] = R[i] - L[i];
    }
    m -= s;
    if (m < 0) return;
    for (long long j = 0; j <= m; ++j) DP[0][j] = 1;
    for (long long i = 1; i <= n; ++i) {
      for (long long j = 0; j <= m; ++j) {
        if (j - C[i] - 1 >= 0)
          DP[i][j] = (DP[i - 1][j] - DP[i - 1][j - C[i] - 1]) % 998244353;
        else
          DP[i][j] = DP[i - 1][j];
        if (j) DP[i][j] = (DP[i][j] + DP[i][j - 1]) % 998244353;
      }
    }
    if (!DP[n][m]) {
      DPF[t] = 0;
      return;
    }
    DPF[t] = (DP[n][m] + 998244353) % 998244353;
  }
  signed main() {
    cin >> n >> mm;
    for (long long i = 1; i <= n; ++i) cin >> A[i] >> B[i];
    for (long long i = 1; i <= mm; ++i) solve(i);
    for (long long i = mm; i >= 1; --i) {
      for (long long j = 2 * i; j <= mm; j += i)
        DPF[i] = (DPF[i] - DPF[j] + 998244353) % 998244353;
    }
    cout << DPF[1] << "\n";
    return 0;
  }
  