#include <bits/stdc++.h>
using namespace std;
long long read() {
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  long long num = 0;
  while (isdigit(c)) num = num * 10 + c - '0', c = getchar();
  return num;
}
vector<long long> prime;
int inv[100];
int fact[100], invfact[100];
int main() {
  long long d = read();
  for (int i = 2; 1ll * i * i <= d; i++)
    if (d % i == 0) {
      prime.push_back(i);
      while (d % i == 0) d /= i;
    }
  if (d > 1) prime.push_back(d);
  inv[1] = 1;
  for (int i = 2; i < 100; i++)
    inv[i] = 1ll * (998244353 - 998244353 / i) * inv[998244353 % i] % 998244353;
  fact[0] = invfact[0] = 1;
  for (int i = 1; i < 100; i++)
    fact[i] = 1ll * fact[i - 1] * i % 998244353,
    invfact[i] = 1ll * invfact[i - 1] * inv[i] % 998244353;
  int q = read();
  for (int i = 1; i <= q; i++) {
    long long a, b;
    a = read(), b = read();
    int a1 = 0, a2 = 0, ans = 1;
    for (long long i : prime) {
      int c1 = 0, c2 = 0;
      while (a % i == 0) ++c1, a /= i;
      while (b % i == 0) ++c2, b /= i;
      if (c1 == c2) continue;
      if (c1 > c2)
        a1 += c1 - c2;
      else
        a2 += c2 - c1;
      ans = 1ll * ans * invfact[abs(c1 - c2)] % 998244353;
    }
    cout << 1ll * ans * fact[a1] % 998244353 * fact[a2] % 998244353 << endl;
  }
}

### 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;
long long read() {
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  long long num = 0;
  while (isdigit(c)) num = num * 10 + c - '0', c = getchar();
  return num;
}
vector<long long> prime;
int inv[100];
int fact[100], invfact[100];
int main() {
  long long d = read();
  for (int i = 2; 1ll * i * i <= d; i++)
    if (d % i == 0) {
      prime.push_back(i);
      while (d % i == 0) d /= i;
    }
  if (d > 1) prime.push_back(d);
  inv[1] = 1;
  for (int i = 2; i < 100; i++)
    inv[i] = 1ll * (998244353 - 998244353 / i) * inv[998244353 % i] % 998244353;
  fact[0] = invfact[0] = 1;
  for (int i = 1; i < 100; i++)
    fact[i] = 1ll * fact[i - 1] * i % 998244353,
    invfact[i] = 1ll * invfact[i - 1] * inv[i] % 998244353;
  int q = read();
  for (int i = 1; i <= q; i++) {
    long long a, b;
    a = read(), b = read();
    int a1 = 0, a2 = 0, ans = 1;
    for (long long i : prime) {
      int c1 = 0, c2 = 0;
      while (a % i == 0) ++c1, a /= i;
      while (b % i == 0) ++c2, b /= i;
      if (c1 == c2) continue;
      if (c1 > c2)
        a1 += c1 - c2;
      else
        a2 += c2 - c1;
      ans = 1ll * ans * invfact[abs(c1 - c2)] % 998244353;
    }
    cout << 1ll * ans * fact[a1] % 998244353 * fact[a2] % 998244353 << endl;
  }
}

```