#include <bits/stdc++.h>
using namespace std;
int n, q;
long long u, v, d;
long long ft[101], rft[101];
std::vector<long long> primes, powe;
long long power(long long b, long long p) {
  long long res = 1, a = b;
  for (; p; p >>= 1) {
    if (p & 1) res = res * a % 998244353;
    a = a * a % 998244353;
  }
  return res;
}
void build() {
  for (long long i = 2; i * i <= d; ++i) {
    if (d % i == 0) {
      primes.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) primes.push_back(d);
  powe.resize(primes.size());
  ft[0] = ft[1] = 1;
  for (int i = 2; i <= 100; ++i) ft[i] = ft[i - 1] * i % 998244353;
  rft[100] = power(ft[100], 998244353 - 2);
  for (int i = 99; i > 0; --i) rft[i] = rft[i + 1] * (i + 1) % 998244353;
}
void quer() {
  cin >> u >> v;
  for (int i = 0; i < primes.size(); ++i) {
    powe[i] = 0;
    long long k = primes[i];
    while (u % k == 0) {
      u /= k;
      ++powe[i];
    }
    while (v % k == 0) {
      v /= k;
      --powe[i];
    }
  }
  std::sort(powe.begin(), powe.end());
  long long ans = 1;
  int c = 0;
  for (int i = 0; i < primes.size() && powe[i] < 0; ++i) {
    c += -powe[i];
    ans = ans * rft[-powe[i]] % 998244353;
  }
  ans = ans * ft[c] % 998244353;
  c = 0;
  for (int i = primes.size() - 1; i >= 0 && powe[i] > 0; --i) {
    c += powe[i];
    ans = ans * rft[powe[i]] % 998244353;
  }
  ans = ans * ft[c] % 998244353;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> d >> q;
  build();
  for (int i = 0; i < q; ++i) quer();
  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;
int n, q;
long long u, v, d;
long long ft[101], rft[101];
std::vector<long long> primes, powe;
long long power(long long b, long long p) {
  long long res = 1, a = b;
  for (; p; p >>= 1) {
    if (p & 1) res = res * a % 998244353;
    a = a * a % 998244353;
  }
  return res;
}
void build() {
  for (long long i = 2; i * i <= d; ++i) {
    if (d % i == 0) {
      primes.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) primes.push_back(d);
  powe.resize(primes.size());
  ft[0] = ft[1] = 1;
  for (int i = 2; i <= 100; ++i) ft[i] = ft[i - 1] * i % 998244353;
  rft[100] = power(ft[100], 998244353 - 2);
  for (int i = 99; i > 0; --i) rft[i] = rft[i + 1] * (i + 1) % 998244353;
}
void quer() {
  cin >> u >> v;
  for (int i = 0; i < primes.size(); ++i) {
    powe[i] = 0;
    long long k = primes[i];
    while (u % k == 0) {
      u /= k;
      ++powe[i];
    }
    while (v % k == 0) {
      v /= k;
      --powe[i];
    }
  }
  std::sort(powe.begin(), powe.end());
  long long ans = 1;
  int c = 0;
  for (int i = 0; i < primes.size() && powe[i] < 0; ++i) {
    c += -powe[i];
    ans = ans * rft[-powe[i]] % 998244353;
  }
  ans = ans * ft[c] % 998244353;
  c = 0;
  for (int i = primes.size() - 1; i >= 0 && powe[i] > 0; --i) {
    c += powe[i];
    ans = ans * rft[powe[i]] % 998244353;
  }
  ans = ans * ft[c] % 998244353;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> d >> q;
  build();
  for (int i = 0; i < q; ++i) quer();
  return 0;
}

```