#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
long long inv[maxn], p[maxn], s[maxn];
long long C(long long n, long long m) {
  return p[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
}
int main() {
  long long t;
  t = 1;
  while (t--) {
    long long d, cnt = 0;
    scanf("%lld", &d);
    for (long long i = 2; i * i <= d; i++) {
      if (d % i == 0) s[++cnt] = i;
      while (d % i == 0) d /= i;
    }
    if (d != 1ll) s[++cnt] = d;
    inv[0] = inv[1] = p[0] = p[1] = 1ll;
    for (long long i = 2; i <= 102; i++) {
      p[i] = p[i - 1] * i % 998244353;
      inv[i] = (998244353 - (998244353 / i)) * inv[998244353 % i] % 998244353;
    }
    for (long long i = 2; i <= 102; i++)
      inv[i] = inv[i] * inv[i - 1] % 998244353;
    long long m;
    scanf("%lld", &m);
    while (m--) {
      long long ans = 1, sx = 0, sy = 0, x, y;
      scanf("%lld %lld", &x, &y);
      for (long long i = 1; i <= cnt; i++) {
        long long xa = 0, yb = 0, mx = 0;
        while (x % s[i] == 0) {
          x /= s[i];
          xa++;
        }
        while (y % s[i] == 0) {
          y /= s[i];
          yb++;
        }
        mx = max(xa, yb);
        sx += mx - xa;
        sy += mx - yb;
        ans = ans * C(sx, mx - xa) % 998244353 * C(sy, mx - yb) % 998244353;
      }
      printf("%lld\n", ans % 998244353);
    }
  }
}

### 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 namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
long long inv[maxn], p[maxn], s[maxn];
long long C(long long n, long long m) {
  return p[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
}
int main() {
  long long t;
  t = 1;
  while (t--) {
    long long d, cnt = 0;
    scanf("%lld", &d);
    for (long long i = 2; i * i <= d; i++) {
      if (d % i == 0) s[++cnt] = i;
      while (d % i == 0) d /= i;
    }
    if (d != 1ll) s[++cnt] = d;
    inv[0] = inv[1] = p[0] = p[1] = 1ll;
    for (long long i = 2; i <= 102; i++) {
      p[i] = p[i - 1] * i % 998244353;
      inv[i] = (998244353 - (998244353 / i)) * inv[998244353 % i] % 998244353;
    }
    for (long long i = 2; i <= 102; i++)
      inv[i] = inv[i] * inv[i - 1] % 998244353;
    long long m;
    scanf("%lld", &m);
    while (m--) {
      long long ans = 1, sx = 0, sy = 0, x, y;
      scanf("%lld %lld", &x, &y);
      for (long long i = 1; i <= cnt; i++) {
        long long xa = 0, yb = 0, mx = 0;
        while (x % s[i] == 0) {
          x /= s[i];
          xa++;
        }
        while (y % s[i] == 0) {
          y /= s[i];
          yb++;
        }
        mx = max(xa, yb);
        sx += mx - xa;
        sy += mx - yb;
        ans = ans * C(sx, mx - xa) % 998244353 * C(sy, mx - yb) % 998244353;
      }
      printf("%lld\n", ans % 998244353);
    }
  }
}

```