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 int const rem = 998244353;
vector<long long int> prime_factors;
vector<long long int> factorial;
vector<long long int> inverse_fac;
long long int modInverse(long long int a) {
  long long int m = rem;
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
void compute() {
  factorial.resize(100);
  inverse_fac.resize(100);
  factorial[0] = 1;
  for (long long int i = 1; i < 100; i++)
    factorial[i] = (factorial[i - 1] * i) % rem;
  for (long long int i = 0; i < 100; i++)
    inverse_fac[i] = modInverse(factorial[i]);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void primeFactors(long long int n) {
  if (n % 2 == 0) prime_factors.push_back(2);
  while (n % 2 == 0) n = n / 2;
  for (long long int i = 3; i * i <= n; i = i + 2) {
    if (n % i == 0) {
      prime_factors.push_back(i);
      while (n % i == 0) n = n / i;
    }
  }
  if (n > 2) prime_factors.push_back(n);
}
long long int comb(long long int n, long long int r) {
  long long int answer = factorial[n];
  answer = (answer * inverse_fac[n - r]) % rem;
  answer = (answer * inverse_fac[r]) % rem;
  return answer;
}
long long int count(long long int n) {
  long long int answer = 1;
  vector<long long int> a;
  long long int cnt = 0;
  long long int total = 0;
  for (long long int i : prime_factors) {
    cnt = 0;
    while (n % i == 0) {
      n = n / i;
      cnt++;
    }
    total += cnt;
    if (cnt > 0) a.push_back(cnt);
  }
  for (long long int i : a) {
    answer = (answer * comb(total, i)) % rem;
    total -= i;
  }
  return answer;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  compute();
  long long int d;
  cin >> d;
  primeFactors(d);
  long long int q;
  cin >> q;
  while (q--) {
    long long int u, v;
    cin >> u >> v;
    long long int w = gcd(u, v);
    u = u / w;
    v = v / w;
    long long int facu = count(u);
    long long int facv = count(v);
    cout << (facu * facv) % rem << '\n';
  }
}