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 maxn = 4e5 + 10;
const long long mod = 998244353;
const long long inf = 1e18 + 10;
const long long Log = 21;
long long n, m, k;
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  long long x = pw(a, b / 2);
  x = (x * x) % mod;
  if (b & 1) x = (x * a) % mod;
  return x;
}
long long cnt(long long x, long long p) {
  long long X = 0;
  while (x % p == 0) {
    x /= p;
    X++;
  }
  return X;
}
long long D;
vector<long long> pr;
long long fac[maxn], repo[maxn];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fac[0] = 1;
  repo[0] = 1;
  for (long long i = 1; i < 200; i++) {
    fac[i] = (fac[i - 1] * i) % mod;
    repo[i] = pw(fac[i], mod - 2);
  }
  cin >> D;
  for (long long i = 2; i <= sqrt(D); i++)
    if (D % i == 0) {
      while (D % i == 0) D /= i;
      pr.push_back(i);
    }
  if (D > 1) pr.push_back(D);
  long long q;
  cin >> q;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    long long ans = 1, sm2 = 0, sm1 = 0;
    for (auto p : pr) {
      long long x = cnt(u, p);
      long long y = cnt(v, p);
      (ans *= repo[abs(x - y)]) %= mod;
      if (x < y) sm1 += y - x;
      if (x > y) sm2 += x - y;
    }
    (ans *= (fac[sm1] * fac[sm2]) % mod) %= mod;
    cout << ans << "\n";
  }
}