#include <bits/stdc++.h>
using namespace std;
long long int fact[51];
long long int pw(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int val = pw(a, n / 2);
  val = (val * val) % 998244353;
  if (n % 2) val = (val * a) % 998244353;
  return val;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int D;
  cin >> D;
  fact[0] = 1;
  for (long long int i = 1; i <= 50; i++)
    fact[i] = (fact[i - 1] * i) % 998244353;
  vector<long long int> div;
  long long int num = D;
  for (long long int i = 2; i * i <= num; i++) {
    if (num % i == 0) div.push_back(i);
    while (num % i == 0) num = num / i;
  }
  if (num != 1) div.push_back(num);
  long long int leng = div.size();
  long long int q, v, u;
  cin >> q;
  while (q--) {
    cin >> v >> u;
    vector<long long int> vp(leng, 0), up(leng, 0), lcm(leng, 0);
    long long int dn = 1;
    long long int u_num = 0;
    long long int v_num = 0;
    for (long long int i = 0; i < leng; i++) {
      long long int no = div[i];
      while (v % no == 0) {
        v /= no;
        vp[i] = (vp[i] + 1) % 998244353;
      }
      while (u % no == 0) {
        u /= no;
        up[i] = (up[i] + 1) % 998244353;
      }
      lcm[i] = max(vp[i], up[i]);
      u_num = (u_num + (lcm[i] - up[i])) % 998244353;
      v_num = (v_num + (lcm[i] - vp[i])) % 998244353;
      dn = (dn * fact[lcm[i] - up[i]]) % 998244353;
      dn = (dn * fact[lcm[i] - vp[i]]) % 998244353;
    }
    long long int ans = (fact[u_num] * fact[v_num]) % 998244353;
    ans = (ans * pw(dn, 998244353 - 2)) % 998244353;
    cout << ans << '\n';
  }
}

### Prompt

Your task is to create a cpp solution to the following problem:

You are given a positive integer D. Let's build the following graph from it: 

  * each vertex is a divisor of D (not necessarily prime, 1 and D itself are also included); 
  * two vertices x and y (x > y) have an undirected edge between them if x is divisible by y and \frac x y is a prime; 
  * the weight of an edge is the number of divisors of x that are not divisors of y. 



For example, here is the graph for D=12: 

<image>

Edge (4,12) has weight 3 because 12 has divisors [1,2,3,4,6,12] and 4 has divisors [1,2,4]. Thus, there are 3 divisors of 12 that are not divisors of 4 — [3,6,12].

There is no edge between 3 and 2 because 3 is not divisible by 2. There is no edge between 12 and 3 because 12/3=4 is not a prime.

Let the length of the path between some vertices v and u in the graph be the total weight of edges on it. For example, path [(1, 2), (2, 6), (6, 12), (12, 4), (4, 2), (2, 6)] has length 1+2+2+3+1+2=11. The empty path has length 0.

So the shortest path between two vertices v and u is the path that has the minimal possible length.

Two paths a and b are different if there is either a different number of edges in them or there is a position i such that a_i and b_i are different edges.

You are given q queries of the following form: 

  * v u — calculate the number of the shortest paths between vertices v and u. 



The answer for each query might be large so print it modulo 998244353.

Input

The first line contains a single integer D (1 ≤ D ≤ 10^{15}) — the number the graph is built from.

The second line contains a single integer q (1 ≤ q ≤ 3 ⋅ 10^5) — the number of queries.

Each of the next q lines contains two integers v and u (1 ≤ v, u ≤ D). It is guaranteed that D is divisible by both v and u (both v and u are divisors of D).

Output

Print q integers — for each query output the number of the shortest paths between the two given vertices modulo 998244353.

Examples

Input


12
3
4 4
12 1
3 4


Output


1
3
1


Input


1
1
1 1


Output


1


Input


288807105787200
4
46 482955026400
12556830686400 897
414 12556830686400
4443186242880 325


Output


547558588
277147129
457421435
702277623

Note

In the first example: 

  * The first query is only the empty path — length 0; 
  * The second query are paths [(12, 4), (4, 2), (2, 1)] (length 3+1+1=5), [(12, 6), (6, 2), (2, 1)] (length 2+2+1=5) and [(12, 6), (6, 3), (3, 1)] (length 2+2+1=5). 
  * The third query is only the path [(3, 1), (1, 2), (2, 4)] (length 1+1+1=3). 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int fact[51];
long long int pw(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int val = pw(a, n / 2);
  val = (val * val) % 998244353;
  if (n % 2) val = (val * a) % 998244353;
  return val;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int D;
  cin >> D;
  fact[0] = 1;
  for (long long int i = 1; i <= 50; i++)
    fact[i] = (fact[i - 1] * i) % 998244353;
  vector<long long int> div;
  long long int num = D;
  for (long long int i = 2; i * i <= num; i++) {
    if (num % i == 0) div.push_back(i);
    while (num % i == 0) num = num / i;
  }
  if (num != 1) div.push_back(num);
  long long int leng = div.size();
  long long int q, v, u;
  cin >> q;
  while (q--) {
    cin >> v >> u;
    vector<long long int> vp(leng, 0), up(leng, 0), lcm(leng, 0);
    long long int dn = 1;
    long long int u_num = 0;
    long long int v_num = 0;
    for (long long int i = 0; i < leng; i++) {
      long long int no = div[i];
      while (v % no == 0) {
        v /= no;
        vp[i] = (vp[i] + 1) % 998244353;
      }
      while (u % no == 0) {
        u /= no;
        up[i] = (up[i] + 1) % 998244353;
      }
      lcm[i] = max(vp[i], up[i]);
      u_num = (u_num + (lcm[i] - up[i])) % 998244353;
      v_num = (v_num + (lcm[i] - vp[i])) % 998244353;
      dn = (dn * fact[lcm[i] - up[i]]) % 998244353;
      dn = (dn * fact[lcm[i] - vp[i]]) % 998244353;
    }
    long long int ans = (fact[u_num] * fact[v_num]) % 998244353;
    ans = (ans * pw(dn, 998244353 - 2)) % 998244353;
    cout << ans << '\n';
  }
}

```