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;
  using lli = long long int;
  using ulli = unsigned long long int;
  using Ld = long double;
  using pii = pair<int, int>;
  using pll = pair<lli, lli>;
  using pld = pair<Ld, Ld>;
  int pi[100001];
  const lli MOD = 998244353;
  pii p[51];
  lli dp[51][100001];
  lli GetRes(int n, int m, int g) {
    for (int i = (0); i <= (m); i++) dp[0][i] = 1;
    for (int k = (0); k < (n); k++) {
      lli l = (p[k].first + g - 1) / g, r = p[k].second / g;
      if (l > r) return 0;
      for (int i = (1); i <= (m); i++) {
        lli s = 0;
        if (i - l >= 0) s = dp[k][i - l];
        lli mi = 0;
        if (i - r - 1 >= 0) mi = dp[k][i - r - 1];
        dp[k + 1][i] = s - mi;
        dp[k + 1][i] = (dp[k + 1][i] + MOD) % MOD;
      }
      dp[k + 1][0] = 0;
      for (int i = (1); i <= (m); i++) {
        dp[k + 1][i] += dp[k + 1][i - 1];
        dp[k + 1][i] %= MOD;
      }
    }
    return dp[n][m];
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = (0); i < (n); i++) cin >> p[i].first >> p[i].second;
    pi[1] = 1;
    lli res = 0;
    for (int i = (1); i <= (m / n + 1); i++) {
      if (pi[i] == 0) continue;
      res += pi[i] * GetRes(n, m / i, i);
      res = (res + MOD) % MOD;
      for (int j = i + i; j <= m / n + 1; j += i) pi[j] -= pi[i];
    }
    cout << res;
    cout.flush();
    return 0;
  }
  