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 long long int N = 100001;
const long long int M = 1;
struct point {
  long long int x, y;
  long long int i;
};
long long int fast_exp(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n & 1)
    return fast_exp(x, n - 1) * x % 998244353;
  else {
    long long int tmp = fast_exp(x, n >> 1);
    return tmp * tmp % 998244353;
  }
}
vector<long long int> p;
void Sieve_of_Eratosthenes(long long int n) {
  bool *Primes = new bool[n + 1];
  Primes[0] = Primes[1] = 0;
  for (long long int i = 2; i < n + 1; i++) Primes[i] = 1;
  for (long long int i = 2; i * i <= n; i++) {
    if (Primes[i] == 1) {
      for (long long int j = i; i * j <= n; j++) {
        Primes[i * j] = 0;
      }
    }
  }
  for (long long int i = 2; i < n + 1; i++) {
    if (Primes[i]) {
      p.push_back(i);
    }
  }
  delete[] Primes;
}
long long int fact[1001], infact[1001];
map<long long int, long long int> mp;
long long int solve(long long int x) {
  if (mp.find(x) != mp.end()) return mp[x];
  long long int xx = x;
  int n = p.size();
  long long int res = 1, sum = 0;
  for (int i = 0; i < n && p[i] * p[i] <= x; i++) {
    if (x % p[i] == 0) {
      long long int cnt = 0;
      while (x % p[i] == 0) {
        x /= p[i];
        cnt++;
      }
      res = res * infact[cnt] % 998244353;
      sum += cnt;
    }
  }
  if (x > 1) sum++;
  return mp[xx] = res * fact[sum] % 998244353;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0 || b == 0) return a + b;
  return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int d;
  cin >> d;
  Sieve_of_Eratosthenes(sqrt(d) + 1);
  int _ = 1;
  cin >> _;
  fact[0] = fact[1] = 1;
  infact[0] = infact[1] = 1;
  for (long long int i = 2; i < 1001; i++) {
    fact[i] = fact[i - 1] * (i) % 998244353;
    infact[i] = fast_exp(fact[i], 998244353 - 2);
  }
  while (_--) {
    long long int n, m, x = 0, y, ans = 0, l, r;
    cin >> x >> y;
    l = gcd(x, y);
    l = x / l * y;
    x = l / x;
    y = l / y;
    ans = solve(x) * solve(y) % 998244353;
    cout << ans << "\n";
  }
  return 0;
}