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 = 1e18;
  const long double pi = 3.141592653589793238;
  const long long MOD = 998244353;
  const long long N = 2e5 + 10;
  long long n, m;
  pair<long long, long long> a[51];
  long long dp[51][N];
  long long ans[N] = {0};
  long long pre[N];
  long long sum(long long l, long long r) {
    if (r < 0) return 0;
    if (l > r) return 0;
    return pre[r] - (l - 1 >= 0 ? pre[l - 1] : 0);
  }
  void solve() {
    cin >> n >> m;
    for (long long i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
    for (long long g = m; g >= 1; g--) {
      long long mx = m / g;
      for (long long i = 0; i <= n; i++) {
        for (long long j = 0; j <= mx; j++) {
          dp[i][j] = 0;
        }
      }
      dp[0][0] = 1;
      for (long long i = 1; i <= n; i++) {
        long long l = (a[i - 1].first + g - 1) / g, r = a[i - 1].second / g;
        pre[0] = dp[i - 1][0];
        for (long long j = 1; j <= mx; j++)
          pre[j] = dp[i - 1][j] + pre[j - 1], pre[j] %= MOD;
        for (long long j = 0; j <= mx; j++) {
          long long nl = j - r;
          long long nr = j - l;
          dp[i][j] = sum(nl, nr);
          dp[i][j] %= MOD;
          dp[i][j] += MOD;
          dp[i][j] %= MOD;
        }
      }
      for (long long j = 1; j <= mx; j++) ans[g] += dp[n][j], ans[g] %= MOD;
      for (long long k = 2; k * g <= m; k++) {
        ans[g] -= ans[k * g];
        ans[g] %= MOD;
        ans[g] += MOD;
        ans[g] %= MOD;
      }
    }
    cout << ans[1];
  }
  signed main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    long long t = 1;
    while (t--) {
      solve();
    }
  }
  