#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long mul(long long a, long long b) { return a * b % mod; }
long long binpow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = mul(ans, a);
    }
    b >>= 1;
    a = mul(a, a);
  }
  return ans;
}
long long div(long long a) { return binpow(a, mod - 2); }
long long fact[10000];
long long precalk[200];
signed main() {
  fact[0] = 1;
  for (long long i = 1; i < 10000; i++) {
    fact[i] = mul(i, fact[i - 1]);
  }
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long d;
  cin >> d;
  long long sqrd = 2;
  for (; sqrd * sqrd <= d; sqrd++) {
  }
  vector<long long> p;
  for (long long i = 2; i <= sqrd; i++) {
    if (d % i == 0) {
      p.push_back(i);
      while (d % i == 0) {
        d /= i;
      }
    }
  }
  if (d != 1) {
    p.push_back(d);
  }
  long long n = p.size();
  for (long long i = 0; i < 200; i++) {
    precalk[i] = div(fact[i]);
  }
  long long q;
  cin >> q;
  long long v, u;
  for (long long c = 0; c < q; c++) {
    cin >> v >> u;
    vector<long long> p1(n), p2(n);
    for (long long j = 0; j < n; j++) {
      while (v % p[j] == 0) {
        p1[j]++;
        v /= p[j];
      }
      while (u % p[j] == 0) {
        p2[j]++;
        u /= p[j];
      }
    }
    long long s1 = 0;
    long long s2 = 0;
    vector<long long> v1(n, 0), v2(n, 0);
    for (long long i = 0; i < n; i++) {
      if (p1[i] > p2[i]) {
        s1 += p1[i] - p2[i];
        v1[i] = p1[i] - p2[i];
      } else {
        s2 += p2[i] - p1[i];
        v2[i] = p2[i] - p1[i];
      }
    }
    long long ans = mul(fact[s1], fact[s2]);
    for (long long i = 0; i < n; i++) {
      if (v1[i] != 0) {
        ans = mul(ans, precalk[v1[i]]);
      }
      if (v2[i] != 0) {
        ans = mul(ans, precalk[v2[i]]);
      }
    }
    cout << ans << '\n';
  }
}

### Prompt

Your task is to create a CPP solution to the following 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). 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long mul(long long a, long long b) { return a * b % mod; }
long long binpow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = mul(ans, a);
    }
    b >>= 1;
    a = mul(a, a);
  }
  return ans;
}
long long div(long long a) { return binpow(a, mod - 2); }
long long fact[10000];
long long precalk[200];
signed main() {
  fact[0] = 1;
  for (long long i = 1; i < 10000; i++) {
    fact[i] = mul(i, fact[i - 1]);
  }
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long d;
  cin >> d;
  long long sqrd = 2;
  for (; sqrd * sqrd <= d; sqrd++) {
  }
  vector<long long> p;
  for (long long i = 2; i <= sqrd; i++) {
    if (d % i == 0) {
      p.push_back(i);
      while (d % i == 0) {
        d /= i;
      }
    }
  }
  if (d != 1) {
    p.push_back(d);
  }
  long long n = p.size();
  for (long long i = 0; i < 200; i++) {
    precalk[i] = div(fact[i]);
  }
  long long q;
  cin >> q;
  long long v, u;
  for (long long c = 0; c < q; c++) {
    cin >> v >> u;
    vector<long long> p1(n), p2(n);
    for (long long j = 0; j < n; j++) {
      while (v % p[j] == 0) {
        p1[j]++;
        v /= p[j];
      }
      while (u % p[j] == 0) {
        p2[j]++;
        u /= p[j];
      }
    }
    long long s1 = 0;
    long long s2 = 0;
    vector<long long> v1(n, 0), v2(n, 0);
    for (long long i = 0; i < n; i++) {
      if (p1[i] > p2[i]) {
        s1 += p1[i] - p2[i];
        v1[i] = p1[i] - p2[i];
      } else {
        s2 += p2[i] - p1[i];
        v2[i] = p2[i] - p1[i];
      }
    }
    long long ans = mul(fact[s1], fact[s2]);
    for (long long i = 0; i < n; i++) {
      if (v1[i] != 0) {
        ans = mul(ans, precalk[v1[i]]);
      }
      if (v2[i] != 0) {
        ans = mul(ans, precalk[v2[i]]);
      }
    }
    cout << ans << '\n';
  }
}

```