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 int MOD = 998244353;
const int MAX = 4E7;
const int MAXF = 1E3;
int sieve[MAX];
int fac[MAXF];
int invFac[MAXF];
int modinv(int x) {
  int ans = 1;
  int k = MOD - 2;
  while (k) {
    if (k & 1) {
      ans = (int)(((long long)ans * x) % MOD);
    }
    k >>= 1;
    x = (int)(((long long)x * x) % MOD);
  }
  return ans;
}
int choose(int n, int k) {
  int ans = (int)(((long long)fac[n] * invFac[k]) % MOD);
  return (int)(((long long)ans * invFac[n - k]) % MOD);
}
int calc(multiset<long long> &s1, multiset<long long> &s2) {
  long long prev = -1;
  int ans = 1;
  int cnt = 0;
  for (auto it = s1.begin(); it != s1.end(); it++) {
    if (*it == prev) continue;
    prev = *it;
    long long target = *it;
    int diff = s1.count(target) - s2.count(target);
    if (diff > 0) {
      cnt += diff;
      ans = (int)(((long long)ans * choose(cnt, diff)) % MOD);
    }
  }
  return ans;
}
int main() {
  fac[0] = invFac[0] = 1;
  for (int i = 1; i < MAXF; i++) {
    fac[i] = (int)(((long long)fac[i - 1] * i) % MOD);
    invFac[i] = modinv(fac[i]);
  }
  vector<int> primes;
  for (int i = 2; i < MAX; i++) {
    if (sieve[i] == 0) {
      sieve[i] = i;
      primes.push_back(i);
      if ((long long)i * i < MAX)
        for (int j = i * i; j < MAX; j += i) sieve[j] = i;
    }
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long d;
  cin >> d;
  int q;
  cin >> q;
  multiset<long long> div;
  vector<long long> divv;
  for (int x : primes) {
    if (d % x == 0) divv.push_back(x);
    while (d % x == 0) {
      d /= x;
      div.insert(x);
    }
  }
  if (d > 1) {
    div.insert(d);
    divv.push_back(d);
  }
  while (q--) {
    long long v, u;
    cin >> v >> u;
    multiset<long long> s1, s2;
    for (long long x : divv) {
      while (v % x == 0) {
        v /= x;
        s1.insert(x);
      }
      while (u % x == 0) {
        u /= x;
        s2.insert(x);
      }
    }
    int ans1 = calc(s1, s2);
    int ans2 = calc(s2, s1);
    int ans = (int)(((long long)ans1 * ans2) % MOD);
    cout << ans << "\n";
  }
  return 0;
}