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 int maxn = 110000;
  const long long modi = 998244353;
  long long f[maxn], fact[maxn], dp[50][maxn], pre[50][maxn];
  int l[maxn], r[maxn], l2[maxn], r2[maxn], mu[maxn];
  bool prime[maxn];
  vector<int> p;
  long long quickpow(long long base, long long times) {
    long long ans = 1;
    while (times) {
      if (times & 1) ans = ans * base % modi;
      base = base * base % modi;
      times >>= 1;
    }
    return ans;
  }
  long long inv(long long x) { return quickpow(x, modi - 2); }
  int upper_div(int x, int d) {
    if (x % d == 0) return x / d;
    return x / d + 1;
  }
  inline long long get_pre(int i, int l, int r) {
    if (l > 0) return ((pre[i][r] - pre[i][l - 1]) % modi + modi) % modi;
    return pre[i][r];
  }
  long long tot(int n, int m) {
    for (int i = 0; i <= m; i++) {
      for (int j = 0; j <= n; j++) {
        dp[j][i] = pre[j][i] = 0;
      }
    }
    dp[0][0] = 1;
    for (int i = 0; i <= m; i++) {
      pre[0][i] = 1;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= m; j++) {
        dp[i][j] = get_pre(i - 1, j - r2[i], j);
      }
      pre[i][0] = dp[i][0];
      for (int j = 1; j <= m; j++) {
        pre[i][j] = (pre[i][j - 1] + dp[i][j]) % modi;
      }
    }
    return pre[n][m];
  }
  int main() {
    int n, m;
    cin >> n >> m;
    fact[0] = 1;
    for (int i = 1; i <= n + m; i++) {
      fact[i] = fact[i - 1] * i % modi;
    }
    mu[1] = 1;
    for (int i = 1; i <= n; i++) {
      cin >> l[i] >> r[i];
    }
    for (int i = 2; i <= m; i++) {
      if (prime[i] == false) {
        mu[i] = -1;
        p.push_back(i);
      }
      for (auto v : p) {
        if (i * v > m) break;
        mu[i * v] = mu[i] * (-1);
        prime[i * v] = true;
        if (i % v == 0) {
          mu[i * v] = 0;
          break;
        }
      }
    }
    long long ret = 0;
    for (int d = 1; d <= m; d++) {
      int m2 = m / d;
      bool tf = true;
      for (int i = 1; i <= n; i++) {
        l2[i] = upper_div(l[i], d);
        m2 -= l2[i];
        r2[i] = r[i] / d;
        if (l2[i] > r2[i]) {
          tf = false;
          break;
        }
        r2[i] -= l2[i];
      }
      if (tf && m2 >= 0) {
        ret = (ret + mu[d] * tot(n, m2)) % modi;
        ret = (ret + modi) % modi;
      }
    }
    cout << ret << endl;
  }
  