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 N = 1e6 + 100;
long long int mod = 998244353;
long long int rev[N], fac[N];
vector<long long int> v;
map<pair<long long int, long long int>, long long int> mp;
long long int wop(long long int x, long long int y) {
  long long int ret = 1;
  for (; y >= 1; y /= 2) {
    if (y & 1) ret *= x;
    x *= x;
    x %= mod;
    ret %= mod;
  }
  return ret;
}
long long int chos(long long int r, long long int n) {
  long long int ret = fac[n] * rev[r];
  ret %= mod;
  ret *= rev[n - r];
  ret %= mod;
  return ret;
}
long long int how(vector<long long int> &p) {
  long long int sum = 0;
  for (int u : p) sum += u;
  long long int ret = 1;
  for (int u : p) {
    ret *= chos(u, sum);
    sum -= u;
    ret %= mod;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  fac[0] = 1;
  for (int i = 1; i <= 1e5; i++) fac[i] = (fac[i - 1] * i) % mod;
  for (int i = 0; i <= 1e5; i++) rev[i] = wop(fac[i], mod - 2);
  long long int n;
  cin >> n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      v.emplace_back(i);
      while (n % i == 0) n /= i;
    }
  }
  if (n != 1) v.emplace_back(n);
  long long int q;
  cin >> q;
  while (q--) {
    long long int a, b;
    cin >> a >> b;
    long long int a1 = a, b1 = b;
    if (a == b) {
      cout << 1 << '\n';
      continue;
    }
    if (mp[make_pair(a, b)] != 0) {
      cout << mp[make_pair(a, b)] << '\n';
      continue;
    }
    vector<long long int> de, in;
    de.clear();
    in.clear();
    for (int i = 0; i < v.size(); i++) {
      if (a % v[i] == 0 || b % v[i] == 0) {
        long long int dif = 0;
        while (a % v[i] == 0 && b % v[i] == 0) {
          a /= v[i];
          b /= v[i];
        }
        while (a % v[i] == 0) {
          a /= v[i];
          dif++;
        }
        if (dif != 0)
          de.emplace_back(dif);
        else {
          while (b % v[i] == 0) {
            b /= v[i];
            dif++;
          }
          if (dif != 0) in.emplace_back(dif);
        }
      }
    }
    long long int ans = how(in) * how(de);
    ans %= mod;
    mp[make_pair(a1, b1)] = mp[make_pair(b1, a1)] = ans;
    cout << ans << '\n';
  }
}