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 INF_LL = 1e18;
  const int INF = 1e9 + 100;
  const int MOD = 998244353;
  const long double EPS = 1e-10;
  const int BLOCK = 550;
  const int dx[4] = {-1, 0, 1, 0};
  const int dy[4] = {0, 1, 0, -1};
  mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
  void open_file() {}
  const int maxN = 6e5 + 100;
  void sol() {
    int n, m;
    cin >> n >> m;
    vector<int> L(n), R(n);
    for (int i = 0; i < n; i++) cin >> L[i] >> R[i];
    auto calc = [&](int x) {
      for (int i = 0; i < n; i++) {
        int a = (L[i] + x - 1) / x;
        int b = R[i] / x;
        if (a > b || b == 0) return 0;
      }
      int nm = m / x;
      vector<int> dp(nm + 1, 0);
      dp[0] = 1;
      for (int i = 0; i < n; i++) {
        int a = (L[i] + x - 1) / x;
        int b = R[i] / x;
        if (a > b) return 0;
        vector<int> n_dp(nm + 1, 0);
        for (int j = 0; j + a <= nm; j++) {
          int na = j + a;
          int nb = j + b;
          n_dp[na] = (n_dp[na] + dp[j]) % MOD;
          if (nb < nm) n_dp[nb + 1] = (n_dp[nb + 1] - dp[j] + MOD) % MOD;
        }
        swap(n_dp, dp);
        for (int i = 1; i <= nm; i++) dp[i] = (dp[i - 1] + dp[i]) % MOD;
        if (*max_element(dp.begin(), dp.end()) == 0) return 0;
      }
      int ans = 0;
      for (int i = 0; i <= nm; i++) ans = (ans + dp[i]) % MOD;
      return ans;
    };
    vector<int> F(m + 1);
    for (int i = 1; i <= m; i++) F[i] = calc(i);
    for (int i = m; i >= 1; i--) {
      for (int j = i + i; j <= m; j += i) F[i] = (F[i] - F[j] + MOD) % MOD;
    }
    cout << F[1];
  }
  void solve() {
    int T = 1;
    int TestCase = 0;
    while (T--) {
      TestCase += 1;
      sol();
    }
  }
  int main(int argc, char *argv[]) {
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    open_file();
    solve();
    return 0;
  }
  