#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2")
using namespace std;
using lld = long long int;
using pii = pair<int, int>;
using pll = pair<lld, lld>;
lld n;
lld gcd(lld p, lld q) {
  while (p % q) {
    lld t = p % q;
    p = q;
    q = t;
  }
  return q;
}
lld p[100009];
int pl;
lld fact[100009];
lld inv[100009];
lld invf[100009];
lld cal(lld x) {
  lld res = 1;
  int cnt = 0;
  for (int i = 0; i < pl && p[i] <= x; i++) {
    int j = 0;
    while (x % p[i] == 0) {
      x /= p[i];
      j++;
      cnt++;
    }
    res = (res * invf[j]) % 998244353;
  }
  res = (res * fact[cnt]) % 998244353;
  return res;
}
void inil() {
  lld z = n;
  lld i;
  pl = 0;
  for (i = 2; i <= z / i; i++) {
    if (z % i == 0) {
      p[pl++] = i;
      while (z % i == 0) {
        z /= i;
      }
    }
  }
  if (z > 1) {
    p[pl++] = z;
  }
  fact[0] = fact[1] = 1;
  inv[0] = inv[1] = 1;
  invf[0] = invf[1] = 1;
  for (i = 2; i <= 100000; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
    inv[i] = (998244353 - (998244353 / i * inv[998244353 % i]) % 998244353) %
             998244353;
    invf[i] = (invf[i - 1] * inv[i]) % 998244353;
  }
}
int main() {
  int i, j, k;
  int t = 1, tv = 0;
  scanf("%lld", &n);
  inil();
  scanf("%d", &t);
  while (t--) {
    lld p, q;
    lld g;
    scanf("%lld %lld", &p, &q);
    g = gcd(p, q);
    if (p == q) {
      printf("1\n");
      continue;
    }
    printf("%lld\n", (cal(p / g) * cal(q / g)) % 998244353);
  }
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2")
using namespace std;
using lld = long long int;
using pii = pair<int, int>;
using pll = pair<lld, lld>;
lld n;
lld gcd(lld p, lld q) {
  while (p % q) {
    lld t = p % q;
    p = q;
    q = t;
  }
  return q;
}
lld p[100009];
int pl;
lld fact[100009];
lld inv[100009];
lld invf[100009];
lld cal(lld x) {
  lld res = 1;
  int cnt = 0;
  for (int i = 0; i < pl && p[i] <= x; i++) {
    int j = 0;
    while (x % p[i] == 0) {
      x /= p[i];
      j++;
      cnt++;
    }
    res = (res * invf[j]) % 998244353;
  }
  res = (res * fact[cnt]) % 998244353;
  return res;
}
void inil() {
  lld z = n;
  lld i;
  pl = 0;
  for (i = 2; i <= z / i; i++) {
    if (z % i == 0) {
      p[pl++] = i;
      while (z % i == 0) {
        z /= i;
      }
    }
  }
  if (z > 1) {
    p[pl++] = z;
  }
  fact[0] = fact[1] = 1;
  inv[0] = inv[1] = 1;
  invf[0] = invf[1] = 1;
  for (i = 2; i <= 100000; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
    inv[i] = (998244353 - (998244353 / i * inv[998244353 % i]) % 998244353) %
             998244353;
    invf[i] = (invf[i - 1] * inv[i]) % 998244353;
  }
}
int main() {
  int i, j, k;
  int t = 1, tv = 0;
  scanf("%lld", &n);
  inil();
  scanf("%d", &t);
  while (t--) {
    lld p, q;
    lld g;
    scanf("%lld %lld", &p, &q);
    g = gcd(p, q);
    if (p == q) {
      printf("1\n");
      continue;
    }
    printf("%lld\n", (cal(p / g) * cal(q / g)) % 998244353);
  }
}

```