#include <bits/stdc++.h>
using ll = long long;
const int MOD = 998244353;
const int MC = 500;
int Q, c[MC][MC];
ll D, u, v;
std::vector<ll> pr;
ll solve(std::vector<int>& a) {
  ll f = 1;
  int l = 0;
  for (int x : a) f = f * c[l += x][x] % MOD;
  return f;
}
std::vector<int> a, b;
int main(void) {
  for (int i = 0; i < MC; ++i) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; ++j) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
  }
  scanf("%lld", &D);
  for (int i = 2; (ll)i * i <= D; ++i) {
    if (D % i == 0) pr.push_back(i);
    while (D % i == 0) D /= i;
  }
  if (D > 1) pr.push_back(D);
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    scanf("%lld%lld", &u, &v);
    for (ll j : pr) {
      int x = 0, y = 0;
      while (u % j == 0) u /= j, ++x;
      while (v % j == 0) v /= j, ++y;
      if (x > y)
        a.push_back(x - y);
      else
        b.push_back(y - x);
    }
    printf("%lld\n", solve(a) * solve(b) % MOD);
    a.clear(), b.clear();
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 ll = long long;
const int MOD = 998244353;
const int MC = 500;
int Q, c[MC][MC];
ll D, u, v;
std::vector<ll> pr;
ll solve(std::vector<int>& a) {
  ll f = 1;
  int l = 0;
  for (int x : a) f = f * c[l += x][x] % MOD;
  return f;
}
std::vector<int> a, b;
int main(void) {
  for (int i = 0; i < MC; ++i) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; ++j) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
  }
  scanf("%lld", &D);
  for (int i = 2; (ll)i * i <= D; ++i) {
    if (D % i == 0) pr.push_back(i);
    while (D % i == 0) D /= i;
  }
  if (D > 1) pr.push_back(D);
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    scanf("%lld%lld", &u, &v);
    for (ll j : pr) {
      int x = 0, y = 0;
      while (u % j == 0) u /= j, ++x;
      while (v % j == 0) v /= j, ++y;
      if (x > y)
        a.push_back(x - y);
      else
        b.push_back(y - x);
    }
    printf("%lld\n", solve(a) * solve(b) % MOD);
    a.clear(), b.clear();
  }
  return 0;
}

```