#include <bits/stdc++.h>
using namespace std;
long long n, d, q, u, v, mod = 998244353ll, invfact[100], fact[100], cont, res,
                         tot, tot2;
vector<pair<long long, int> > fatt;
long long fastp(long long base, long long exp) {
  if (exp == 0) return 1ll;
  long long tmp = fastp(base, exp / 2);
  tmp = (tmp * tmp) % mod;
  if (exp % 2 == 1) tmp *= base;
  return tmp % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fact[0] = 1;
  for (long long i = 1; i < 100; i++) fact[i] = (fact[i - 1] * i) % mod;
  invfact[99] = fastp(fact[99], mod - 2ll);
  for (long long i = 98; i >= 0; i--)
    invfact[i] = (invfact[i + 1] * (i + 1ll)) % mod;
  cin >> d;
  for (long long i = 2; (i * i) <= d; i++) {
    if (d % i == 0) {
      cont = 0;
      while (d % i == 0) {
        cont++;
        d /= i;
      }
      fatt.push_back({i, cont});
    }
  }
  if (d != 1) fatt.push_back({d, 1});
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> u >> v;
    if (u < v) swap(u, v);
    tot = tot2 = 0;
    res = 1;
    for (int j = 0; j < fatt.size(); j++) {
      cont = 0;
      while (u % fatt[j].first == 0) {
        u /= fatt[j].first;
        cont++;
      }
      while (v % fatt[j].first == 0) {
        v /= fatt[j].first;
        cont--;
      }
      res *= invfact[abs(cont)];
      res %= mod;
      if (cont > 0)
        tot += cont;
      else
        tot2 += -cont;
    }
    res *= fact[tot];
    res %= mod;
    res *= fact[tot2];
    res %= mod;
    cout << res << "\n";
  }
  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;
long long n, d, q, u, v, mod = 998244353ll, invfact[100], fact[100], cont, res,
                         tot, tot2;
vector<pair<long long, int> > fatt;
long long fastp(long long base, long long exp) {
  if (exp == 0) return 1ll;
  long long tmp = fastp(base, exp / 2);
  tmp = (tmp * tmp) % mod;
  if (exp % 2 == 1) tmp *= base;
  return tmp % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fact[0] = 1;
  for (long long i = 1; i < 100; i++) fact[i] = (fact[i - 1] * i) % mod;
  invfact[99] = fastp(fact[99], mod - 2ll);
  for (long long i = 98; i >= 0; i--)
    invfact[i] = (invfact[i + 1] * (i + 1ll)) % mod;
  cin >> d;
  for (long long i = 2; (i * i) <= d; i++) {
    if (d % i == 0) {
      cont = 0;
      while (d % i == 0) {
        cont++;
        d /= i;
      }
      fatt.push_back({i, cont});
    }
  }
  if (d != 1) fatt.push_back({d, 1});
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> u >> v;
    if (u < v) swap(u, v);
    tot = tot2 = 0;
    res = 1;
    for (int j = 0; j < fatt.size(); j++) {
      cont = 0;
      while (u % fatt[j].first == 0) {
        u /= fatt[j].first;
        cont++;
      }
      while (v % fatt[j].first == 0) {
        v /= fatt[j].first;
        cont--;
      }
      res *= invfact[abs(cont)];
      res %= mod;
      if (cont > 0)
        tot += cont;
      else
        tot2 += -cont;
    }
    res *= fact[tot];
    res %= mod;
    res *= fact[tot2];
    res %= mod;
    cout << res << "\n";
  }
  return 0;
}

```