#include <bits/stdc++.h>
const int M = 3e5 + 10;
using namespace std;
int Q, top;
int h[M], inv[M], ST[M];
long long B, D, st[M];
map<long long, int> MP;
int Pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % 998244353;
    b >>= 1;
    a = 1ll * a * a % 998244353;
  }
  return ans;
}
void Pre() {
  h[0] = inv[0] = 1;
  for (int i = 1; i <= 2e5; i++) {
    h[i] = 1ll * h[i - 1] * i % 998244353;
    inv[i] = Pow(h[i], 998244353 - 2);
  }
}
int C(int n, int m) {
  return 1ll * h[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
}
long long GCD(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return GCD(b, a % b);
}
int Cal(long long x) {
  int sum = 0, tot = 0, ans = 1;
  for (int i = 1; i <= top; i++)
    if (x % st[i] == 0) {
      ++tot;
      while (x % st[i] == 0) ++ST[tot], x /= st[i];
    }
  for (int i = 1; i <= tot; i++) sum += ST[i];
  for (int i = 1; i <= tot; i++) {
    ans = 1ll * ans * C(sum, ST[i]) % 998244353;
    sum -= ST[i];
    ST[i] = 0;
  }
  return ans;
}
int main() {
  scanf("%lld%d", &D, &Q);
  B = D;
  for (long long i = 2; i * i <= D; i++)
    if (B % i == 0) {
      st[++top] = i;
      while (B % i == 0) B /= i;
    }
  if (B != 1) st[++top] = B;
  Pre();
  while (Q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    long long gc = GCD(x, y);
    int t = Cal(x / gc);
    int m = Cal(y / gc);
    if (!MP[x / gc]) MP[x / gc] = t;
    if (!MP[y / gc]) MP[y / gc] = m;
    printf("%d\n", 1ll * MP[x / gc] * MP[y / gc] % 998244353);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
const int M = 3e5 + 10;
using namespace std;
int Q, top;
int h[M], inv[M], ST[M];
long long B, D, st[M];
map<long long, int> MP;
int Pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % 998244353;
    b >>= 1;
    a = 1ll * a * a % 998244353;
  }
  return ans;
}
void Pre() {
  h[0] = inv[0] = 1;
  for (int i = 1; i <= 2e5; i++) {
    h[i] = 1ll * h[i - 1] * i % 998244353;
    inv[i] = Pow(h[i], 998244353 - 2);
  }
}
int C(int n, int m) {
  return 1ll * h[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
}
long long GCD(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return GCD(b, a % b);
}
int Cal(long long x) {
  int sum = 0, tot = 0, ans = 1;
  for (int i = 1; i <= top; i++)
    if (x % st[i] == 0) {
      ++tot;
      while (x % st[i] == 0) ++ST[tot], x /= st[i];
    }
  for (int i = 1; i <= tot; i++) sum += ST[i];
  for (int i = 1; i <= tot; i++) {
    ans = 1ll * ans * C(sum, ST[i]) % 998244353;
    sum -= ST[i];
    ST[i] = 0;
  }
  return ans;
}
int main() {
  scanf("%lld%d", &D, &Q);
  B = D;
  for (long long i = 2; i * i <= D; i++)
    if (B % i == 0) {
      st[++top] = i;
      while (B % i == 0) B /= i;
    }
  if (B != 1) st[++top] = B;
  Pre();
  while (Q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    long long gc = GCD(x, y);
    int t = Cal(x / gc);
    int m = Cal(y / gc);
    if (!MP[x / gc]) MP[x / gc] = t;
    if (!MP[y / gc]) MP[y / gc] = m;
    printf("%d\n", 1ll * MP[x / gc] * MP[y / gc] % 998244353);
  }
  return 0;
}

```