#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353, N = 1e5 + 5, M = 1e5 + 5, INF = 0x3f3f3f3f;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int MAX = 66;
long long finv[MAX], inv[MAX], f[MAX];
void init() {
  finv[0] = finv[1] = f[0] = f[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i < MAX; ++i) {
    f[i] = i * f[i - 1] % mod;
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    finv[i] = finv[i - 1] * inv[i] % mod;
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long D;
  cin >> D;
  init();
  int q;
  cin >> q;
  vector<long long> p;
  for (long long i = 2; i * i <= D; ++i) {
    if (D % i == 0) {
      p.push_back(i);
      D /= i;
      while (D % i == 0) D /= i;
    }
  }
  if (D > 1) p.push_back(D);
  while (q--) {
    long long u, v, g, sum1 = 0, sum2 = 0, ans1 = 1, ans2 = 1;
    cin >> u >> v;
    g = gcd(u, v);
    u /= g;
    v /= g;
    for (auto i : p) {
      int num = 0;
      while (u % i == 0) {
        ++sum1;
        u /= i;
        ++num;
      }
      ans1 = ans1 * finv[num] % mod;
    }
    ans1 = ans1 * f[sum1] % mod;
    for (auto i : p) {
      int num = 0;
      while (v % i == 0) {
        ++sum2;
        v /= i;
        ++num;
      }
      ans2 = ans2 * finv[num] % mod;
    }
    ans2 = ans2 * f[sum2] % mod;
    cout << ans1 * ans2 % mod << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, N = 1e5 + 5, M = 1e5 + 5, INF = 0x3f3f3f3f;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int MAX = 66;
long long finv[MAX], inv[MAX], f[MAX];
void init() {
  finv[0] = finv[1] = f[0] = f[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i < MAX; ++i) {
    f[i] = i * f[i - 1] % mod;
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    finv[i] = finv[i - 1] * inv[i] % mod;
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long D;
  cin >> D;
  init();
  int q;
  cin >> q;
  vector<long long> p;
  for (long long i = 2; i * i <= D; ++i) {
    if (D % i == 0) {
      p.push_back(i);
      D /= i;
      while (D % i == 0) D /= i;
    }
  }
  if (D > 1) p.push_back(D);
  while (q--) {
    long long u, v, g, sum1 = 0, sum2 = 0, ans1 = 1, ans2 = 1;
    cin >> u >> v;
    g = gcd(u, v);
    u /= g;
    v /= g;
    for (auto i : p) {
      int num = 0;
      while (u % i == 0) {
        ++sum1;
        u /= i;
        ++num;
      }
      ans1 = ans1 * finv[num] % mod;
    }
    ans1 = ans1 * f[sum1] % mod;
    for (auto i : p) {
      int num = 0;
      while (v % i == 0) {
        ++sum2;
        v /= i;
        ++num;
      }
      ans2 = ans2 * finv[num] % mod;
    }
    ans2 = ans2 * f[sum2] % mod;
    cout << ans1 * ans2 % mod << endl;
  }
  return 0;
}

```