#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
  return x % y == 0 ? y : gcd(y, x % y);
}
long long fat[123456], mod = 998244353;
bool composite[40000000];
vector<long long> prime;
long long modPow(long long b, long long e) {
  if (e == 0) return 1;
  if (e == 1) return b % mod;
  long long res = modPow(b, e / 2);
  res = (res * res) % mod;
  if (e % 2 != 0) res = (res * (b % mod)) % mod;
  return res;
}
long long solve(long long u) {
  long long res = 1;
  long long p = 0, q = 1;
  for (long long pr : prime) {
    if (pr * pr > u) break;
    long long cnt = 0;
    while (u % pr == 0) {
      cnt++;
      u /= pr;
      ++p;
    }
    if (cnt) {
      q = (q * fat[cnt]) % mod;
    }
  }
  if (u != 1) ++p;
  return (fat[p] * modPow(q, mod - 2)) % mod;
}
int main() {
  long long k;
  cin >> k;
  long long q;
  cin >> q;
  long long root = sqrt(k) + 2;
  for (long long i = 2; i <= root; i++)
    if (!composite[i]) {
      if (k % i == 0) prime.push_back(i);
      for (long long j = i * i; j <= root; j += i) composite[j] = 1;
    }
  fat[0] = 1;
  for (int i = 1; i < 100000; ++i) fat[i] = (fat[i - 1] * i) % mod;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    long long g = gcd(u, v);
    u /= g;
    v /= g;
    long long r1 = solve(u);
    long long r2 = solve(v);
    long long res = (r1 * r2) % mod;
    cout << res << 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;
long long gcd(long long x, long long y) {
  return x % y == 0 ? y : gcd(y, x % y);
}
long long fat[123456], mod = 998244353;
bool composite[40000000];
vector<long long> prime;
long long modPow(long long b, long long e) {
  if (e == 0) return 1;
  if (e == 1) return b % mod;
  long long res = modPow(b, e / 2);
  res = (res * res) % mod;
  if (e % 2 != 0) res = (res * (b % mod)) % mod;
  return res;
}
long long solve(long long u) {
  long long res = 1;
  long long p = 0, q = 1;
  for (long long pr : prime) {
    if (pr * pr > u) break;
    long long cnt = 0;
    while (u % pr == 0) {
      cnt++;
      u /= pr;
      ++p;
    }
    if (cnt) {
      q = (q * fat[cnt]) % mod;
    }
  }
  if (u != 1) ++p;
  return (fat[p] * modPow(q, mod - 2)) % mod;
}
int main() {
  long long k;
  cin >> k;
  long long q;
  cin >> q;
  long long root = sqrt(k) + 2;
  for (long long i = 2; i <= root; i++)
    if (!composite[i]) {
      if (k % i == 0) prime.push_back(i);
      for (long long j = i * i; j <= root; j += i) composite[j] = 1;
    }
  fat[0] = 1;
  for (int i = 1; i < 100000; ++i) fat[i] = (fat[i - 1] * i) % mod;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    long long g = gcd(u, v);
    u /= g;
    v /= g;
    long long r1 = solve(u);
    long long r2 = solve(v);
    long long res = (r1 * r2) % mod;
    cout << res << endl;
  }
  return 0;
}

```