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;
  int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
  int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
  void file() {}
  void fast() {
    std::ios_base::sync_with_stdio(0);
    cin.tie(NULL);
  }
  int main() {
    file();
    fast();
    int n, m, mod = 998244353;
    cin >> n >> m;
    vector<pair<int, int> > v1(n);
    for (auto& it : v1) cin >> it.first >> it.second;
    auto get_div = [&](pair<int, int> p, int num) {
      return p.second / num - ((p.first - 1) / num);
    };
    auto solve = [&](int g, int sum) -> int {
      vector<pair<int, int> > tmp;
      for (auto& it : v1) {
        if (!get_div(it, g)) return 0;
        pair<int, int> x;
        x.first = (it.first - 1) / g + 1;
        x.second = x.first + get_div(it, g) - 1;
        tmp.push_back(x);
      }
      if (tmp.empty()) return 0;
      vector<int> dp(sum + 1, 1);
      for (auto& it : tmp) {
        vector<int> cur(sum + 1);
        for (int i = 1; i <= sum; i++) {
          int r = max(-1, i - it.first);
          int l = max(-1, i - it.second);
          cur[i] = (r < 0 ? 0 : dp[r]) - (l <= 0 ? 0 : dp[l - 1]);
          cur[i] = (cur[i] + mod) % mod;
        }
        for (int i = 1; i <= sum; i++) cur[i] = (cur[i] + cur[i - 1]) % mod;
        dp = cur;
      }
      return dp[sum];
    };
    vector<int> arr(m + 1, 1);
    for (int i = 1; i <= m; i++) arr[i] = solve(i, m / i);
    for (int i = m; i >= 1; i--)
      for (int j = i + i; j <= m; j += i) arr[i] = (arr[i] - arr[j] + mod) % mod;
    cout << arr[1] << '\n';
    return 0;
  }
  