#include <bits/stdc++.h>
using namespace std;
const int N = 4e7 + 5;
const int mod = 998244353;
int _w;
bool vis[N];
int prime[N / 10], prime_tot, cnt;
long long D, u, v, p[55];
int n, q, A[55], B[55], sum[61], chk[61], fac[61], C[55], ifac[61], inv[61];
void analyze(long long now, int* f) {
  int res;
  for (int i = 1; i <= cnt; ++i) {
    res = 0;
    while (now % p[i] == 0) {
      ++res;
      now /= p[i];
    }
    f[i] = res;
  }
}
int main(void) {
  for (int i = 2; i < N; ++i) {
    if (!vis[i]) prime[++prime_tot] = i;
    for (int j = 1; j <= prime_tot; ++j) {
      if (1LL * i * prime[j] >= N) break;
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  fac[0] = fac[1] = inv[1] = ifac[1] = ifac[0] = 1;
  for (int i = 2; i < 61; ++i) {
    fac[i] = 1LL * fac[i - 1] * i % mod;
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
    ifac[i] = 1LL * ifac[i - 1] * inv[i] % mod;
  }
  _w = scanf("%lld", &D);
  for (int i = 1; i <= prime_tot; ++i) {
    if (D < 2) break;
    if (D % prime[i] == 0) {
      p[++cnt] = prime[i];
      while (D % prime[i] == 0) D /= prime[i];
    }
  }
  if (D > 1) p[++cnt] = D;
  _w = scanf("%d", &n);
  while (n--) {
    _w = scanf("%lld%lld", &u, &v);
    analyze(u, A);
    analyze(v, B);
    int resa = 1, resb = 1, suma = 0, sumb = 0;
    for (int i = 1; i <= cnt; ++i)
      if (A[i] < B[i]) {
        sumb += B[i] - A[i];
        resb = 1LL * resb * ifac[B[i] - A[i]] % mod;
      } else if (A[i] > B[i]) {
        suma += A[i] - B[i];
        resa = 1LL * resa * ifac[A[i] - B[i]] % mod;
      }
    resa = 1LL * resa * fac[suma] % mod;
    resb = 1LL * resb * fac[sumb] % mod;
    printf("%lld\n", 1LL * resa * resb % mod);
  }
  return 0;
}

### Prompt

Generate 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;
const int N = 4e7 + 5;
const int mod = 998244353;
int _w;
bool vis[N];
int prime[N / 10], prime_tot, cnt;
long long D, u, v, p[55];
int n, q, A[55], B[55], sum[61], chk[61], fac[61], C[55], ifac[61], inv[61];
void analyze(long long now, int* f) {
  int res;
  for (int i = 1; i <= cnt; ++i) {
    res = 0;
    while (now % p[i] == 0) {
      ++res;
      now /= p[i];
    }
    f[i] = res;
  }
}
int main(void) {
  for (int i = 2; i < N; ++i) {
    if (!vis[i]) prime[++prime_tot] = i;
    for (int j = 1; j <= prime_tot; ++j) {
      if (1LL * i * prime[j] >= N) break;
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  fac[0] = fac[1] = inv[1] = ifac[1] = ifac[0] = 1;
  for (int i = 2; i < 61; ++i) {
    fac[i] = 1LL * fac[i - 1] * i % mod;
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
    ifac[i] = 1LL * ifac[i - 1] * inv[i] % mod;
  }
  _w = scanf("%lld", &D);
  for (int i = 1; i <= prime_tot; ++i) {
    if (D < 2) break;
    if (D % prime[i] == 0) {
      p[++cnt] = prime[i];
      while (D % prime[i] == 0) D /= prime[i];
    }
  }
  if (D > 1) p[++cnt] = D;
  _w = scanf("%d", &n);
  while (n--) {
    _w = scanf("%lld%lld", &u, &v);
    analyze(u, A);
    analyze(v, B);
    int resa = 1, resb = 1, suma = 0, sumb = 0;
    for (int i = 1; i <= cnt; ++i)
      if (A[i] < B[i]) {
        sumb += B[i] - A[i];
        resb = 1LL * resb * ifac[B[i] - A[i]] % mod;
      } else if (A[i] > B[i]) {
        suma += A[i] - B[i];
        resa = 1LL * resa * ifac[A[i] - B[i]] % mod;
      }
    resa = 1LL * resa * fac[suma] % mod;
    resb = 1LL * resb * fac[sumb] % mod;
    printf("%lld\n", 1LL * resa * resb % mod);
  }
  return 0;
}

```