#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
vector<pair<long long, int> > veki;
int bp[40];
long long fakt[53], inv[53];
long long gcd(long long a, long long b) {
  long long t;
  while (a) {
    t = a;
    a = b % a;
    b = t;
  }
  return b;
}
long long inverz(long long first) {
  long long s = first;
  long long ret = 1ll;
  for (int i = 0; i < 32; i++) {
    if (bp[i]) ret = (ret * s) % mod;
    s = (s * s) % mod;
  }
  return ret;
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 2ll; i * i <= n; i++) {
    if (n % i == 0ll) {
      int ai = 0;
      while (n % i == 0) {
        ai++;
        n /= i;
      }
      veki.push_back(make_pair(i, ai));
    }
  }
  if (n > 1) veki.push_back(make_pair(n, 1));
  int q;
  cin >> q;
  fakt[0] = 1ll;
  int br = 0;
  int k = mod - 2;
  while (k) {
    bp[br] = k % 2;
    k /= 2;
    br++;
  }
  for (long long i = 1ll; i < 51; i++) {
    fakt[i] = (fakt[i - 1] * i) % mod;
  }
  for (long long i = 1ll; i < 51; i++) {
    inv[i] = inverz(fakt[i]);
  }
  for (int i = 0; i < q; i++) {
    long long first, second;
    cin >> first >> second;
    long long g = gcd(first, second);
    long long x1 = first / g;
    long long y1 = second / g;
    int brx = 0, bry = 0;
    long long sx = 1ll, sy = 1ll;
    for (int j = 0; j < veki.size(); j++) {
      if (x1 % veki[j].first == 0ll) {
        int ax = 0;
        while (x1 % veki[j].first == 0ll) {
          ax++;
          brx++;
          x1 /= veki[j].first;
        }
        sx = (sx * inv[ax]) % mod;
      }
      if (y1 % veki[j].first == 0ll) {
        int ay = 0;
        while (y1 % veki[j].first == 0ll) {
          ay++;
          bry++;
          y1 /= veki[j].first;
        }
        sy = (sy * inv[ay]) % mod;
      }
    }
    sx = (sx * fakt[brx]) % mod;
    sy = (sy * fakt[bry]) % mod;
    cout << (sx * sy) % mod << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 int mod = 998244353;
vector<pair<long long, int> > veki;
int bp[40];
long long fakt[53], inv[53];
long long gcd(long long a, long long b) {
  long long t;
  while (a) {
    t = a;
    a = b % a;
    b = t;
  }
  return b;
}
long long inverz(long long first) {
  long long s = first;
  long long ret = 1ll;
  for (int i = 0; i < 32; i++) {
    if (bp[i]) ret = (ret * s) % mod;
    s = (s * s) % mod;
  }
  return ret;
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 2ll; i * i <= n; i++) {
    if (n % i == 0ll) {
      int ai = 0;
      while (n % i == 0) {
        ai++;
        n /= i;
      }
      veki.push_back(make_pair(i, ai));
    }
  }
  if (n > 1) veki.push_back(make_pair(n, 1));
  int q;
  cin >> q;
  fakt[0] = 1ll;
  int br = 0;
  int k = mod - 2;
  while (k) {
    bp[br] = k % 2;
    k /= 2;
    br++;
  }
  for (long long i = 1ll; i < 51; i++) {
    fakt[i] = (fakt[i - 1] * i) % mod;
  }
  for (long long i = 1ll; i < 51; i++) {
    inv[i] = inverz(fakt[i]);
  }
  for (int i = 0; i < q; i++) {
    long long first, second;
    cin >> first >> second;
    long long g = gcd(first, second);
    long long x1 = first / g;
    long long y1 = second / g;
    int brx = 0, bry = 0;
    long long sx = 1ll, sy = 1ll;
    for (int j = 0; j < veki.size(); j++) {
      if (x1 % veki[j].first == 0ll) {
        int ax = 0;
        while (x1 % veki[j].first == 0ll) {
          ax++;
          brx++;
          x1 /= veki[j].first;
        }
        sx = (sx * inv[ax]) % mod;
      }
      if (y1 % veki[j].first == 0ll) {
        int ay = 0;
        while (y1 % veki[j].first == 0ll) {
          ay++;
          bry++;
          y1 /= veki[j].first;
        }
        sy = (sy * inv[ay]) % mod;
      }
    }
    sx = (sx * fakt[brx]) % mod;
    sy = (sy * fakt[bry]) % mod;
    cout << (sx * sy) % mod << endl;
  }
  return 0;
}

```