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;
#pragma GCC push_options
#pragma GCC optimize("unroll-loops")
long long power(long long x, long long y) {
  long long res = 1ll;
  x = x % 998244353;
  while (y) {
    if (y % 2 == 1) res = (res * x) % 998244353;
    x = (x * x) % 998244353;
    y = y / 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long fact[100], inv[100];
  fact[0] = inv[0] = 1;
  for (int i = 1; i < 100; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
  }
  inv[99] = power(fact[99], 998244353 - 2);
  for (int i = 98; i >= 0; i--) inv[i] = (inv[i + 1] * (i + 1)) % 998244353;
  long long d;
  scanf("%lld", &d);
  vector<long long> prime_divisor;
  for (long long i = 2; i * i <= d; i++) {
    if (d % i == 0) {
      prime_divisor.emplace_back(i);
      while (d % i == 0) d = d / i;
    }
  }
  if (d > 1) prime_divisor.emplace_back(d);
  long long q;
  scanf("%lld", &q);
  while (q--) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    vector<long long> pos, neg;
    for (long long x : prime_divisor) {
      long long count = 0ll;
      while (a % x == 0) {
        count--;
        a /= x;
      }
      while (b % x == 0) {
        count++;
        b /= x;
      }
      if (count > 0)
        pos.emplace_back(count);
      else if (count < 0)
        neg.emplace_back(-count);
    }
    long long ans = fact[accumulate(pos.begin(), pos.end(), 0)];
    for (long long x : pos) ans = (ans * inv[x]) % 998244353;
    ans = (ans * fact[accumulate(neg.begin(), neg.end(), 0)]) % 998244353;
    for (long long x : neg) ans = (ans * inv[x]) % 998244353;
    printf("%lld\n", ans);
  }
  return 0;
}