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 powmod(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    exponent /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
const int upperlimit = 71;
const int mod = 998244353;
long long fact[upperlimit];
long long invfact[upperlimit];
vector<long long> pdv;
void func(long long n) {
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) pdv.push_back(i);
    while (n % i == 0) n /= i;
  }
  if (n > 1) pdv.push_back(n);
}
int pw(long long n, long long i) {
  int ans = 0;
  while (n % i == 0) {
    ans++;
    n /= i;
  }
  return ans;
}
int main() {
  fact[0] = 1;
  invfact[0] = 1;
  for (int i = 1; i < upperlimit; i++) {
    fact[i] = fact[i - 1];
    fact[i] *= i;
    fact[i] %= mod;
    invfact[i] = powmod(fact[i], mod - 2, mod);
  }
  long long d, u, v;
  int q;
  long long temp = 1;
  int lol = 0;
  scanf("%lld", &d);
  func(d);
  scanf("%d", &q);
  while (q--) {
    scanf("%lld", &u);
    scanf("%lld", &v);
    long long gc = gcd(u, v);
    temp = 1;
    lol = 0;
    for (int i = 0; i < pdv.size(); i++) {
      int gg = abs(pw(gc, pdv[i]) - pw(u, pdv[i]));
      temp *= invfact[gg];
      temp %= mod;
      lol += gg;
    }
    temp *= fact[lol];
    temp %= mod;
    lol = 0;
    for (int i = 0; i < pdv.size(); i++) {
      int gg = abs(pw(gc, pdv[i]) - pw(v, pdv[i]));
      temp *= invfact[gg];
      temp %= mod;
      lol += gg;
    }
    temp *= fact[lol];
    temp %= mod;
    printf("%lld\n", temp);
  }
  return 0;
}