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;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long qpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % 998244353;
    a = a * a % 998244353;
    n >>= 1;
  }
  return res;
}
long long n;
long long fac[205], inv[205];
vector<long long> v;
void init() {
  fac[0] = 1;
  for (int i = 1; i < 205; i++) fac[i] = fac[i - 1] * i % 998244353;
  inv[205 - 1] = qpow(fac[205 - 1], 998244353 - 2);
  for (int i = 205 - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % 998244353;
}
long long calc(long long x) {
  long long s = 0, res = 1;
  for (int i = 0; i < v.size(); i++) {
    if (x % v[i]) continue;
    int cnt = 0;
    while (x % v[i] == 0) x /= v[i], cnt++;
    s += cnt;
    res = res * inv[cnt] % 998244353;
  }
  res = res * fac[s] % 998244353;
  return res;
}
int main() {
  init();
  cin >> n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i) continue;
    while (n % i == 0) n /= i;
    v.push_back(i);
  }
  if (n > 1) v.push_back(n);
  long long x, y, q, g;
  cin >> q;
  while (q--) {
    scanf("%lld%lld", &x, &y);
    g = gcd(x, y), x /= g, y /= g;
    long long ans = calc(x) * calc(y) % 998244353;
    printf("%lld\n", ans);
  }
  return 0;
}