#include <bits/stdc++.h>
using namespace std;
void swap(long long int &a, long long int &b) {
  long long int temp = a;
  a = b;
  b = temp;
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b % 2) ans = (ans * a) % 998244353;
    a = (a * a) % 998244353;
    b = b >> 1;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (a < b) swap(a, b);
  if (!b) return a;
  return gcd(b, a % b);
}
struct node {
  long long int a, b;
};
vector<vector<long long int> > graph;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(NULL));
  long long int d;
  cin >> d;
  long long int fact[100000];
  fact[0] = 1;
  fact[1] = 1;
  for (long long int i = 2; i < 100000; ++i)
    fact[i] = (i * fact[i - 1]) % 998244353;
  long long int invmod[100000];
  for (long long int i = 1; i < 100000; ++i)
    invmod[i] = power(fact[i], 998244353 - 2);
  vector<long long int> vals;
  long long int temp = sqrt(d) + 1000;
  for (long long int i = 2; i < temp; ++i) {
    if (d % i == 0) {
      vals.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d != 1) vals.push_back(d);
  long long int q;
  cin >> q;
  for (long long int i = 0; i < q; ++i) {
    long long int u, v;
    cin >> u >> v;
    long long int gc = gcd(u, v);
    u /= gc;
    v /= gc;
    long long int a1 = 1;
    vector<long long int> v1;
    long long int sum = 0;
    for (auto j : vals) {
      if (u % j == 0) {
        long long int cnt = 0;
        while (u % j == 0) u /= j, cnt++;
        v1.push_back(cnt);
        sum += cnt;
      }
    }
    a1 = (a1 * fact[sum]) % 998244353;
    for (auto j : v1) {
      a1 = (a1 * invmod[j]) % 998244353;
    }
    long long int a2 = 1;
    vector<long long int> v2;
    sum = 0;
    for (auto j : vals) {
      if (v % j == 0) {
        long long int cnt = 0;
        while (v % j == 0) v /= j, cnt++;
        v2.push_back(cnt);
        sum += cnt;
      }
    }
    a2 = (a2 * fact[sum]) % 998244353;
    for (auto j : v2) {
      a2 = (a2 * invmod[j]) % 998244353;
    }
    long long int ans = (a1 * a2) % 998244353;
    cout << ans << "\n";
  }
  return 0;
}

### 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;
void swap(long long int &a, long long int &b) {
  long long int temp = a;
  a = b;
  b = temp;
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b % 2) ans = (ans * a) % 998244353;
    a = (a * a) % 998244353;
    b = b >> 1;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (a < b) swap(a, b);
  if (!b) return a;
  return gcd(b, a % b);
}
struct node {
  long long int a, b;
};
vector<vector<long long int> > graph;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(NULL));
  long long int d;
  cin >> d;
  long long int fact[100000];
  fact[0] = 1;
  fact[1] = 1;
  for (long long int i = 2; i < 100000; ++i)
    fact[i] = (i * fact[i - 1]) % 998244353;
  long long int invmod[100000];
  for (long long int i = 1; i < 100000; ++i)
    invmod[i] = power(fact[i], 998244353 - 2);
  vector<long long int> vals;
  long long int temp = sqrt(d) + 1000;
  for (long long int i = 2; i < temp; ++i) {
    if (d % i == 0) {
      vals.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d != 1) vals.push_back(d);
  long long int q;
  cin >> q;
  for (long long int i = 0; i < q; ++i) {
    long long int u, v;
    cin >> u >> v;
    long long int gc = gcd(u, v);
    u /= gc;
    v /= gc;
    long long int a1 = 1;
    vector<long long int> v1;
    long long int sum = 0;
    for (auto j : vals) {
      if (u % j == 0) {
        long long int cnt = 0;
        while (u % j == 0) u /= j, cnt++;
        v1.push_back(cnt);
        sum += cnt;
      }
    }
    a1 = (a1 * fact[sum]) % 998244353;
    for (auto j : v1) {
      a1 = (a1 * invmod[j]) % 998244353;
    }
    long long int a2 = 1;
    vector<long long int> v2;
    sum = 0;
    for (auto j : vals) {
      if (v % j == 0) {
        long long int cnt = 0;
        while (v % j == 0) v /= j, cnt++;
        v2.push_back(cnt);
        sum += cnt;
      }
    }
    a2 = (a2 * fact[sum]) % 998244353;
    for (auto j : v2) {
      a2 = (a2 * invmod[j]) % 998244353;
    }
    long long int ans = (a1 * a2) % 998244353;
    cout << ans << "\n";
  }
  return 0;
}

```