#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100;
const int MOD = 998244353;
int fact[105];
int inv[105];
int lgput(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = 1LL * ans * a % MOD;
    a = 1LL * a * a % MOD;
  }
  return ans;
}
map<long long, int> ways;
long long gcd(long long a, long long b) {
  while (b) {
    long long r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  long long d;
  scanf("%lld", &d);
  vector<long long> div;
  for (int i = 1; 1LL * i * i <= d; ++i) {
    if (d % i == 0) {
      div.push_back(i);
      if (d / i != i) div.push_back(d / i);
    }
  }
  vector<long long> pr;
  for (int i = 2; 1LL * i * i <= d; ++i) {
    if (d % i == 0) {
      pr.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) pr.push_back(d);
  fact[0] = 1;
  for (int i = 1; i <= MAXN; ++i) fact[i] = 1LL * fact[i - 1] * i % MOD;
  inv[MAXN] = lgput(fact[MAXN], MOD - 2);
  for (int i = MAXN - 1; i >= 0; --i) inv[i] = 1LL * inv[i + 1] * (i + 1) % MOD;
  for (auto it : div) {
    int nr = 1;
    long long aux = it;
    int dd = 0;
    for (auto it1 : pr) {
      int d1 = 0;
      while (aux % it1 == 0) {
        d1++;
        aux /= it1;
      }
      dd += d1;
      nr = 1LL * nr * inv[d1] % MOD;
    }
    nr = 1LL * nr * fact[dd] % MOD;
    ways[it] = nr;
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    long long g = gcd(u, v);
    int ans1 = ways[u / g];
    int ans2 = ways[v / g];
    printf("%d\n", 1LL * ans1 * ans2 % MOD);
  }
  return 0;
}

### 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 MAXN = 100;
const int MOD = 998244353;
int fact[105];
int inv[105];
int lgput(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = 1LL * ans * a % MOD;
    a = 1LL * a * a % MOD;
  }
  return ans;
}
map<long long, int> ways;
long long gcd(long long a, long long b) {
  while (b) {
    long long r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  long long d;
  scanf("%lld", &d);
  vector<long long> div;
  for (int i = 1; 1LL * i * i <= d; ++i) {
    if (d % i == 0) {
      div.push_back(i);
      if (d / i != i) div.push_back(d / i);
    }
  }
  vector<long long> pr;
  for (int i = 2; 1LL * i * i <= d; ++i) {
    if (d % i == 0) {
      pr.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) pr.push_back(d);
  fact[0] = 1;
  for (int i = 1; i <= MAXN; ++i) fact[i] = 1LL * fact[i - 1] * i % MOD;
  inv[MAXN] = lgput(fact[MAXN], MOD - 2);
  for (int i = MAXN - 1; i >= 0; --i) inv[i] = 1LL * inv[i + 1] * (i + 1) % MOD;
  for (auto it : div) {
    int nr = 1;
    long long aux = it;
    int dd = 0;
    for (auto it1 : pr) {
      int d1 = 0;
      while (aux % it1 == 0) {
        d1++;
        aux /= it1;
      }
      dd += d1;
      nr = 1LL * nr * inv[d1] % MOD;
    }
    nr = 1LL * nr * fact[dd] % MOD;
    ways[it] = nr;
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    long long g = gcd(u, v);
    int ans1 = ways[u / g];
    int ans2 = ways[v / g];
    printf("%d\n", 1LL * ans1 * ans2 % MOD);
  }
  return 0;
}

```