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). 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
vector<long long> fact, ifact, invs;
long long combmod(int n, int k, long long MOD, int M = 1e5) {
  if (!fact.size()) {
    fact.resize(M + 1);
    ifact.resize(M + 1);
    invs.resize(M + 1);
    fact[0] = ifact[0] = invs[0] = fact[1] = ifact[1] = invs[1] = 1;
    for (int i = 2; i <= M; i++) {
      fact[i] = fact[i - 1] * i % MOD;
      invs[i] = MOD - MOD / i * invs[MOD % i] % MOD;
      ifact[i] = ifact[i - 1] * invs[i] % MOD;
    }
  }
  return fact[n] * ifact[k] % MOD * ifact[n - k] % MOD;
}
int main(int argc, char** argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(15);
  if (argc == 2 && atoi(argv[1]) == 123456789) {
    freopen("d:\\code\\cpp\\contests\\stdin", "r", stdin);
  }
  combmod(1, 1, MOD, 1e2);
  long long D;
  cin >> D;
  vector<long long> ps;
  long long p = 2;
  while (p * p <= D) {
    int e = 0;
    if (D % p == 0) ps.push_back(p);
    while (D % p == 0) {
      e++;
      D /= p;
    }
    p++;
  }
  if (D > 1) ps.push_back(D);
  int q;
  cin >> q;
  for (int(i) = 0; (i) < (q); (i)++) {
    long long u, v;
    cin >> u >> v;
    int a[100] = {}, b[100] = {};
    for (int k = 0; k < ps.size(); k++) {
      while (u % ps[k] == 0) {
        u /= ps[k];
        a[k]++;
      }
      while (v % ps[k] == 0) {
        v /= ps[k];
        b[k]++;
      }
    }
    long long sol = 1;
    int s1 = 0, s2 = 0;
    for (int k = 0; k < ps.size(); k++) {
      int g = min(a[k], b[k]);
      s1 += a[k] - g;
      s2 += b[k] - g;
      sol *= ifact[a[k] - g];
      sol %= MOD;
      sol *= ifact[b[k] - g];
      sol %= MOD;
    }
    sol *= fact[s1];
    sol %= MOD;
    sol *= fact[s2];
    sol %= MOD;
    cout << sol << endl;
  }
  if (argc == 2 && atoi(argv[1]) == 123456789)
    cout << clock() * 1.0 / CLOCKS_PER_SEC << " sec\n";
  return 0;
}