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;
  mt19937 rnd(51);
  const long long MOD = 998244353;
  long long mul(long long a, long long b) { return (a * b % MOD); }
  long long sub(long long a, long long b) {
    a -= b;
    if (a < 0) a += MOD;
    return a;
  }
  long long add(long long a, long long b) {
    a += b;
    if (a >= MOD) a -= MOD;
    return a;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long n, m;
    cin >> n >> m;
    vector<pair<long long, long long>> v(n);
    vector<long long> ans(m + 1);
    for (long long i = 0; i < n; i++) {
      cin >> v[i].first >> v[i].second;
    }
    for (long long i = m; i >= 1; --i) {
      long long val = (m / i + 1);
      vector<long long> dp(val);
      dp[0] = 1;
      for (long long j = 0; j < n; ++j) {
        long long first = ((v[j].first + i - 1) / i), brk = v[j].second / i;
        vector<long long> dpp(val);
        for (long long k = 1; k < val; k++) {
          dp[k] = add(dp[k], dp[k - 1]);
        }
        for (long long f = 0; f < val; f++) {
          if (f >= first) {
            dpp[f] = dp[f - first];
          }
          if (f - brk - 1 >= 0) {
            dpp[f] = sub(dpp[f], dp[f - brk - 1]);
          }
        }
        dp = dpp;
      }
      for (long long f = i; f <= m; f += i) {
        ans[i] = sub(ans[i], ans[f]);
      }
      for (long long j = 0; j < val; j++) {
        ans[i] = add(ans[i], dp[j]);
      }
    }
    cout << ans[1] << endl;
    return 0;
  }
  