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 int mx = 1e5, MOD = 998244353;
int fact[mx], ifact[mx];
int add(int a, int b) { return (1LL * a + b) % MOD; }
int mult(int a, int b) { return 1LL * a * b % MOD; }
int modpow(int base, int pow) {
  int res = 1;
  while (pow) {
    if (pow % 2) {
      res = mult(res, base);
    }
    base = mult(base, base);
    pow /= 2;
  }
  return res;
}
int inv(int i) { return modpow(i, MOD - 2); }
int ways(vector<int>& a) {
  int res = 1, sum = 0;
  for (int i = 0; i < a.size(); i++) {
    assert(a[i] >= 0);
    res = mult(res, ifact[a[i]]);
    sum += a[i];
  }
  res = mult(res, fact[sum]);
  assert(sum < mx);
  return res;
}
int main() {
  fact[0] = ifact[0] = 1;
  for (int i = 1; i < mx; i++) {
    fact[i] = mult(i, fact[i - 1]);
    ifact[i] = mult(inv(i), ifact[i - 1]);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long d;
  int q;
  cin >> d >> q;
  vector<long long> primes;
  for (long long i = 2; i * i <= d; i++) {
    if (d % i == 0) {
      primes.push_back(i);
      while (d % i == 0) {
        d /= i;
      }
    }
  }
  if (d > 1) {
    primes.push_back(d);
  }
  while (q--) {
    long long u, v;
    cin >> u >> v;
    vector<int> u_to_gcd, gcd_to_v;
    for (long long p : primes) {
      int diff = 0;
      while (u % p == 0) {
        diff++;
        u /= p;
      }
      while (v % p == 0) {
        diff--;
        v /= p;
      }
      if (diff > 0) {
        u_to_gcd.push_back(diff);
      } else {
        gcd_to_v.push_back(-diff);
      }
    }
    int res = mult(ways(u_to_gcd), ways(gcd_to_v));
    cout << res << '\n';
  }
}