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>
  #pragma GCC optimize("Ofast,unroll-loops")
  #pragma GCC optimize("no-stack-protector,fast-math")
  using namespace std;
  constexpr int N = 3e5 + 7;
  constexpr int M = 998244353;
  long long mu[N], flg[N], p[N], tot;
  void getMu() {
    mu[1] = 1;
    for (long long i = 2; i < N; ++i) {
      if (!flg[i]) p[++tot] = i, mu[i] = -1;
      for (long long j = 1; j <= tot && i * p[j] < N; ++j) {
        flg[i * p[j]] = 1;
        if (i % p[j] == 0) {
          mu[i * p[j]] = 0;
          break;
        }
        mu[i * p[j]] = -mu[i];
      }
    }
  }
  long long dp[2][N];
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    getMu();
    long long n, m;
    cin >> n >> m;
    vector<pair<long long, long long> > a(n + 1);
    for (long long i = 1; i <= n; i++) {
      cin >> a[i].first >> a[i].second;
    }
    long long ans = 0;
    for (long long d = 1; d <= m; d++) {
      fill(dp[0], dp[0] + m / d + 1, 1);
      long long x = 0;
      auto sum = [&](long long l, long long r) {
        return ((r >= 0 ? dp[x ^ 1][r] : 0ll) -
                (l - 1 >= 0 ? dp[x ^ 1][l - 1] : 0ll) + M) %
               M;
      };
      for (long long i = 1; i <= n; i++) {
        x ^= 1;
        fill(dp[x], dp[x] + m / d + 1, 0);
        long long l = a[i].first / d + (a[i].first % d != 0);
        long long r = a[i].second / d;
        for (long long j = 0; j <= m / d; j++) {
          dp[x][j] = sum(j - r, j - l);
          if (j - 1 >= 0) dp[x][j] = (dp[x][j] + dp[x][j - 1]) % M;
        }
      }
      ans = (ans + dp[x][m / d] * mu[d] % M + M) % M;
    }
    cout << ans << "\n";
    return 0;
  }
  