#include <bits/stdc++.h>
using namespace std;
long long D;
int Q, n;
long long p[20];
int cnt1[20], cnt2[20], cnt3[20];
long long fact[101], invFact[101];
const long long mod = 998244353;
long long fstPow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ret;
}
int main() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i <= 100; i++) fact[i] = fact[i - 1] * i % mod;
  invFact[100] = fstPow(fact[100], mod - 2);
  for (int i = 99; i >= 0; i--) invFact[i] = invFact[i + 1] * (i + 1) % mod;
  scanf("%lld%d", &D, &Q);
  int sqrtD = sqrt(D);
  long long d = D;
  for (int i = 2; i <= sqrtD && i <= d; i++)
    if (d % i == 0) {
      p[++n] = i;
      while (d % i == 0) d /= i;
    }
  if (d > 1) p[++n] = d;
  long long u, v, Ans;
  int g1, g2;
  for (int i = 1; i <= Q; i++) {
    memset(cnt1, 0, sizeof(cnt1));
    memset(cnt2, 0, sizeof(cnt2));
    g1 = g2 = 0;
    Ans = 1;
    scanf("%lld%lld", &u, &v);
    for (int j = 1; j <= n; j++)
      while (u % p[j] == 0) ++cnt1[j], u /= p[j];
    for (int j = 1; j <= n; j++)
      while (v % p[j] == 0) ++cnt2[j], v /= p[j];
    for (int j = 1; j <= n; j++) {
      cnt3[j] = min(cnt1[j], cnt2[j]);
      g1 += cnt1[j] - cnt3[j];
      g2 += cnt2[j] - cnt3[j];
      Ans = Ans * invFact[cnt1[j] - cnt3[j]] % mod *
            invFact[cnt2[j] - cnt3[j]] % mod;
    }
    Ans = Ans * fact[g1] % mod * fact[g2] % mod;
    printf("%lld\n", Ans);
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
long long D;
int Q, n;
long long p[20];
int cnt1[20], cnt2[20], cnt3[20];
long long fact[101], invFact[101];
const long long mod = 998244353;
long long fstPow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ret;
}
int main() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i <= 100; i++) fact[i] = fact[i - 1] * i % mod;
  invFact[100] = fstPow(fact[100], mod - 2);
  for (int i = 99; i >= 0; i--) invFact[i] = invFact[i + 1] * (i + 1) % mod;
  scanf("%lld%d", &D, &Q);
  int sqrtD = sqrt(D);
  long long d = D;
  for (int i = 2; i <= sqrtD && i <= d; i++)
    if (d % i == 0) {
      p[++n] = i;
      while (d % i == 0) d /= i;
    }
  if (d > 1) p[++n] = d;
  long long u, v, Ans;
  int g1, g2;
  for (int i = 1; i <= Q; i++) {
    memset(cnt1, 0, sizeof(cnt1));
    memset(cnt2, 0, sizeof(cnt2));
    g1 = g2 = 0;
    Ans = 1;
    scanf("%lld%lld", &u, &v);
    for (int j = 1; j <= n; j++)
      while (u % p[j] == 0) ++cnt1[j], u /= p[j];
    for (int j = 1; j <= n; j++)
      while (v % p[j] == 0) ++cnt2[j], v /= p[j];
    for (int j = 1; j <= n; j++) {
      cnt3[j] = min(cnt1[j], cnt2[j]);
      g1 += cnt1[j] - cnt3[j];
      g2 += cnt2[j] - cnt3[j];
      Ans = Ans * invFact[cnt1[j] - cnt3[j]] % mod *
            invFact[cnt2[j] - cnt3[j]] % mod;
    }
    Ans = Ans * fact[g1] % mod * fact[g2] % mod;
    printf("%lld\n", Ans);
  }
  return 0;
}

```