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;
namespace combo {
const long long MOD = 998244353;
vector<long long> inv = {-1, 1};
vector<long long> fact = {1, 1};
vector<long long> ifact = {1, 1};
int precalculated = 1;
void calc_factorials(int n) {
  if (n <= precalculated) {
    return;
  }
  fact.resize(n + 1);
  inv.resize(n + 1);
  ifact.resize(n + 1);
  for (int i = precalculated + 1; i <= n; i++) {
    fact[i] = fact[i - 1] * i % MOD;
    inv[i] = -MOD / i * inv[MOD % i] % MOD;
    ifact[i] = ifact[i - 1] * inv[i] % MOD;
  }
  precalculated = n;
}
long long choose(int n, int k) {
  if (n < k) {
    return 0;
  }
  calc_factorials(n);
  return fact[n] * ifact[k] % MOD * ifact[n - k] % MOD;
}
long long permute(int n, int k) {
  if (n < k) {
    return 0;
  }
  calc_factorials(n);
  return fact[n] * ifact[n - k] % MOD;
}
}  // namespace combo
using namespace combo;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long d;
  cin >> d;
  vector<pair<long long, int>> f;
  long long dd = d;
  for (long long x = 2; x * x <= dd; x++) {
    int e = 0;
    while (d % x == 0) {
      d /= x;
      e++;
    }
    if (e > 0) {
      f.emplace_back(x, e);
    }
  }
  if (d > 1) {
    f.emplace_back(d, 1);
  }
  int q;
  cin >> q;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    vector<pair<int, int>> exps;
    vector<int> diffs;
    for (const auto &p : f) {
      int a = 0;
      while (u % p.first == 0) {
        u /= p.first;
        a++;
      }
      int b = 0;
      while (v % p.first == 0) {
        v /= p.first;
        b++;
      }
      exps.emplace_back(a, b);
      diffs.emplace_back(a - b);
    }
    vector<int> pos, neg;
    for (int d : diffs) {
      if (d > 0) {
        pos.push_back(d);
      }
      if (d < 0) {
        neg.push_back(-d);
      }
    }
    long long ans = 1;
    for (int _ = 0; _ < 2; _++) {
      int sum = accumulate(pos.begin(), pos.end(), 0);
      calc_factorials(sum);
      ans *= fact[sum];
      ans %= MOD;
      for (int d : pos) {
        ans *= ifact[d];
        ans %= MOD;
      }
      swap(pos, neg);
    }
    if (ans < 0) {
      ans += MOD;
    }
    cout << ans << '\n';
  }
  return 0;
}