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 N = 1e5 + 10;
  const long long mod = 998244353;
  int cnt;
  int mo[N];
  bool vis[N];
  int prime[N];
  struct Node {
    int l, r;
  } q[N];
  int n, m;
  void initPrime(const int& n) {
    mo[1] = 1;
    for (int i = 2; i <= n; i++) {
      if (!vis[i]) {
        mo[i] = -1;
        prime[++cnt] = i;
      }
      for (int j = 1; i <= n / prime[j]; j++) {
        vis[i * prime[j]] = true;
        if (i % prime[j] == 0) {
          mo[i * prime[j]] = 0;
          break;
        }
        mo[i * prime[j]] = -mo[i];
      }
    }
  }
  int d[N];
  int pre[N];
  long long get(long long g) {
    vector<Node> vt;
    for (int i = 1; i <= n; i++) {
      int l = (q[i].l + g - 1) / g;
      int r = q[i].r / g;
      if (l > r) return 0;
      vt.push_back({l, r});
    }
    int k = m / g;
    for (int i = vt[0].l; i <= vt[0].r; i++) {
      int l = i + vt[1].l;
      int r = min(k, i + vt[1].r);
      if (l > k) break;
      d[l] += 1, d[r + 1] -= 1;
    }
    for (int i = 1; i <= k; i++) {
      pre[i] = pre[i - 1] + d[i];
      d[i] = 0;
    }
    d[k + 1] = 0;
    for (int i = 2; i <= n - 1; i++) {
      for (int j = 1; j <= k; j++) {
        if (pre[j]) {
          int l = j + vt[i].l;
          int r = min(k, j + vt[i].r);
          if (l > k) break;
          d[l] = (d[l] + pre[j]) % mod;
          d[r + 1] = (d[r + 1] - pre[j] + mod) % mod;
        }
      }
      for (int j = 1; j <= k; j++) {
        pre[j] = (pre[j - 1] + d[j]) % mod;
        d[j] = 0;
      }
      d[k + 1] = 0;
    }
    int res = 0;
    for (int i = 1; i <= k; i++) {
      res = (res + pre[i]) % mod;
    }
    return res;
  }
  int main() {
    initPrime(1e5);
    int mi = 1e5;
    cin >> n >> m;
    for (long long i = 1; i <= n; i++) {
      cin >> q[i].l >> q[i].r;
      mi = min(mi, q[i].r);
    }
    if (n == 1) {
      cout << 1 << endl;
      return 0;
    }
    int ans = 0;
    for (long long i = 1; i <= mi; i++) {
      if (mo[i] == 0) continue;
      ans = (ans + mo[i] * get(i) + mod) % mod;
    }
    cout << ans << endl;
    return 0;
  }
  