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). 
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const long double pi = 3.141592653589793238;
const long long MOD = 998244353;
const long long N = 1e2 + 10;
long long n, fact[N], ifact[N];
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = power(a, b / 2);
  temp *= temp;
  temp %= MOD;
  temp += MOD;
  temp %= MOD;
  if (b % 2) temp *= a;
  temp %= MOD;
  temp += MOD;
  temp %= MOD;
  return temp;
}
void filler() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= MOD;
    fact[i] += MOD;
    fact[i] %= MOD;
  }
  ifact[N - 1] = power(fact[N - 1], MOD - 2);
  for (long long i = N - 2; i >= 0; i--) {
    ifact[i] = (i + 1) * ifact[i + 1];
    ifact[i] %= MOD;
    ifact[i] += MOD;
    ifact[i] %= MOD;
  }
}
void solve() {
  filler();
  cin >> n;
  vector<long long> primes;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i) continue;
    primes.push_back(i);
    while (n % i == 0) n /= i;
  }
  if (n > 1) primes.push_back(n);
  long long q;
  cin >> q;
  for (long long i = 1; i <= q; i++) {
    long long u, v;
    cin >> u >> v;
    vector<long long> up, dn;
    long long dnsum = 0, upsum = 0;
    for (auto x : primes) {
      long long cnt = 0;
      while (u % x == 0) u /= x, cnt--;
      while (v % x == 0) v /= x, cnt++;
      if (cnt < 0)
        dn.push_back(-cnt), dnsum += -cnt;
      else
        up.push_back(cnt), upsum += cnt;
    }
    long long res = 1;
    res *= fact[dnsum];
    for (auto x : dn) res *= ifact[x], res %= MOD, res += MOD, res %= MOD;
    res *= fact[upsum];
    res %= MOD;
    res += MOD;
    res %= MOD;
    for (auto x : up) res *= ifact[x], res %= MOD, res += MOD, res %= MOD;
    cout << res << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}
