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 mod = 998244353;
  int add(int x, int y) {
    x += y;
    if (x >= mod) x -= mod;
    return x;
  }
  int p[2][100001];
  int get(int i, int l, int r) {
    if (r < 0) return 0;
    l = max(l, 0);
    return add(p[i][r], mod - (l == 0 ? 0 : p[i][l - 1]));
  }
  int main() {
    int n, m;
    cin >> n >> m;
    vector<int> f(m + 1);
    vector<int> l(n), r(n);
    for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
    for (int d = 1; d <= m; d++) {
      vector<int> L(n), R(n);
      bool fail = false;
      for (int i = 0; i < n; i++) {
        L[i] = (l[i] + d - 1) / d;
        R[i] = r[i] / d;
        if (R[i] < L[i]) fail = true;
      }
      if (fail) continue;
      int bound = m / d;
      for (int i = 0; i <= 1; i++)
        for (int j = 0; j <= bound; j++) p[i][j] = 0;
      p[0][0] = 1;
      for (int i = 1; i <= bound; i++) p[0][i] = 1;
      int cur = 0, nxt = 1;
      for (int i = 0; i < n; i++) {
        for (int j = 0; j <= bound; j++)
          p[nxt][j] = add(p[nxt][j], get(cur, j - R[i], j - L[i]));
        for (int j = 1; j <= bound; j++)
          p[nxt][j] = add(p[nxt][j], p[nxt][j - 1]);
        for (int j = 0; j <= bound; j++) p[cur][j] = 0;
        swap(cur, nxt);
      }
      f[d] = p[cur][bound];
    }
    for (int i = m; i >= 1; i--) {
      for (int j = i + i; j <= m; j += i) f[i] = add(f[i], mod - f[j]);
    }
    cout << f[1];
  }
  