#include <bits/stdc++.h>
const long long mod = 998244353;
long long factor[105];
long long f[105], invf[105];
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return res;
}
long long inv(long long u) { return power(u, mod - 2); }
void preWork(int n) {
  f[0] = invf[0] = 1;
  for (long long i = 1; i <= n; i++) {
    f[i] = f[i - 1] * i % mod;
    invf[i] = inv(f[i]);
  }
}
int main() {
  preWork(100);
  long long d;
  scanf("%lld", &d);
  int n = 0;
  for (long long i = 2; i * i <= d; i++) {
    if (d == 1) break;
    if (d % i == 0) {
      factor[++n] = i;
      while (d % i == 0) d /= i;
    }
  }
  if (d != 1) factor[++n] = d;
  int q;
  scanf("%d", &q);
  while (q--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    int countP = 0, countN = 0;
    long long divisor = 1;
    for (int i = 1; i <= n; i++) {
      if (u == 1 && v == 1) break;
      long long counta = 0, countb = 0;
      while (u % factor[i] == 0) {
        counta++;
        u /= factor[i];
      }
      while (v % factor[i] == 0) {
        countb++;
        v /= factor[i];
      }
      if (counta != countb) {
        long long diff = counta - countb;
        if (diff > 0) {
          countP += diff;
          divisor = divisor * invf[diff] % mod;
        } else {
          diff *= -1;
          countN += diff;
          divisor = divisor * invf[diff] % mod;
        }
      }
    }
    long long ans = divisor;
    ans = ans * f[countP] % mod;
    ans = ans * f[countN] % mod;
    printf("%lld\n", ans);
  }
  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>
const long long mod = 998244353;
long long factor[105];
long long f[105], invf[105];
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return res;
}
long long inv(long long u) { return power(u, mod - 2); }
void preWork(int n) {
  f[0] = invf[0] = 1;
  for (long long i = 1; i <= n; i++) {
    f[i] = f[i - 1] * i % mod;
    invf[i] = inv(f[i]);
  }
}
int main() {
  preWork(100);
  long long d;
  scanf("%lld", &d);
  int n = 0;
  for (long long i = 2; i * i <= d; i++) {
    if (d == 1) break;
    if (d % i == 0) {
      factor[++n] = i;
      while (d % i == 0) d /= i;
    }
  }
  if (d != 1) factor[++n] = d;
  int q;
  scanf("%d", &q);
  while (q--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    int countP = 0, countN = 0;
    long long divisor = 1;
    for (int i = 1; i <= n; i++) {
      if (u == 1 && v == 1) break;
      long long counta = 0, countb = 0;
      while (u % factor[i] == 0) {
        counta++;
        u /= factor[i];
      }
      while (v % factor[i] == 0) {
        countb++;
        v /= factor[i];
      }
      if (counta != countb) {
        long long diff = counta - countb;
        if (diff > 0) {
          countP += diff;
          divisor = divisor * invf[diff] % mod;
        } else {
          diff *= -1;
          countN += diff;
          divisor = divisor * invf[diff] % mod;
        }
      }
    }
    long long ans = divisor;
    ans = ans * f[countP] % mod;
    ans = ans * f[countN] % mod;
    printf("%lld\n", ans);
  }
  return 0;
}

```