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 ll = long long;
  using ld = long double;
  const int N = 50;
  const int MX = 1e5;
  const int MOD = 998244353;
  int n, m;
  int L[N + 5], R[N + 5];
  int up[N + 5], low[N + 5];
  int f[MX + 5];
  inline void add(int& a, int b) { a = (a + b) % MOD; }
  inline int sub(int a, int b) {
    a = (a - b) % MOD;
    if (a < 0) a += MOD;
    return a;
  }
  int calc(int k) {
    vector<vector<int>> dp(n, vector<int>(m / k + 5));
    vector<int> pre(m / k + 5);
    int bat = m / k;
    for (int i = 0; i < n; i++) {
      low[i] = ((L[i] + k - 1) / k);
      bat -= (low[i]);
      up[i] = ((R[i]) / k);
    }
    if (bat < 0) return 0;
    for (int i = 0; i <= up[0] - low[0]; i++) {
      dp[0][i] = 1;
    }
    pre[0] = 1;
    for (int i = 1; i <= bat; i++) {
      pre[i] = (pre[i - 1] + dp[0][i]) % MOD;
    }
    for (int i = 1; i < n; i++) {
      for (int j = 0; j <= bat; j++) {
        int lul;
        if (j > (up[i] - low[i]))
          lul = pre[j - (up[i] - low[i]) - 1];
        else
          lul = 0;
        dp[i][j] = (dp[i][j] + sub(pre[j], lul)) % MOD;
      }
      pre[0] = dp[i][0];
      for (int j = 1; j <= bat; j++) {
        pre[j] = (pre[j - 1] + dp[i][j]) % MOD;
      }
    }
    int ret = 0;
    for (int i = 0; i <= bat; i++) {
      ret = (ret + dp[n - 1][i]) % MOD;
    }
    return ret;
  }
  void solve() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
      cin >> L[i] >> R[i];
    }
    for (int i = 1; i <= m / n; 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;
      }
    }
    int ans = f[1] % MOD;
    if (ans < 0) ans += MOD;
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int TC = 1;
    for (int i = 1; i <= TC; i++) {
      solve();
    }
    return 0;
  }
  