#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n, m, k, sum1, sum2, tot;
long long s, t, now, ans, a[N], b[N], D;
long long C[1001][1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i <= 1000; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
  }
  cin >> D;
  for (long long i = 2; i * i <= D; ++i) {
    if (D % i == 0) {
      a[++tot] = i;
      while (D % i == 0) D /= i;
    }
  }
  if (D != 1) a[++tot] = D;
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    sum1 = sum2 = 0;
    for (int i = 1; i <= tot; ++i) {
      k = 0;
      while (s % a[i] == 0) {
        ++k;
        s /= a[i];
      }
      while (t % a[i] == 0) {
        --k;
        t /= a[i];
      }
      if (k < 0)
        sum1 += (-k);
      else
        sum2 += k;
      b[i] = k;
    }
    ans = 1;
    for (int i = 1; i <= tot; ++i) {
      if (b[i] < 0) {
        b[i] = -b[i];
        ans = ans * C[sum1][b[i]] % 998244353;
        sum1 -= b[i];
      } else {
        ans = ans * C[sum2][b[i]] % 998244353;
        sum2 -= b[i];
      }
    }
    cout << ans << "\n";
  }
}

### Prompt

In cpp, your task is to solve 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;
const int N = 300005;
int n, m, k, sum1, sum2, tot;
long long s, t, now, ans, a[N], b[N], D;
long long C[1001][1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i <= 1000; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
  }
  cin >> D;
  for (long long i = 2; i * i <= D; ++i) {
    if (D % i == 0) {
      a[++tot] = i;
      while (D % i == 0) D /= i;
    }
  }
  if (D != 1) a[++tot] = D;
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    sum1 = sum2 = 0;
    for (int i = 1; i <= tot; ++i) {
      k = 0;
      while (s % a[i] == 0) {
        ++k;
        s /= a[i];
      }
      while (t % a[i] == 0) {
        --k;
        t /= a[i];
      }
      if (k < 0)
        sum1 += (-k);
      else
        sum2 += k;
      b[i] = k;
    }
    ans = 1;
    for (int i = 1; i <= tot; ++i) {
      if (b[i] < 0) {
        b[i] = -b[i];
        ans = ans * C[sum1][b[i]] % 998244353;
        sum1 -= b[i];
      } else {
        ans = ans * C[sum2][b[i]] % 998244353;
        sum2 -= b[i];
      }
    }
    cout << ans << "\n";
  }
}

```