#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const int md = 998244353;
long long d, x, y, f[20005], of[20005];
vector<long long> a;
bool ff(long long x) {
  long long fff = min((long long)sqrt(x) + 10, x - 1);
  for (long long i = 2; i <= fff; ++i) {
    if (x % i == 0) return 0;
  }
  return 1;
}
long long bin_pow(long long a, int b) {
  long long s = 1;
  while (b > 0) {
    if (b % 2 == 0) {
      a = (1ll * a * a) % md;
      b >>= 1;
    } else {
      s = (1ll * s * a) % md;
      b--;
    }
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> d;
  for (long long i = 2; i <= sqrt(d) + 10; ++i) {
    if (d % i == 0) {
      if (ff(i)) a.push_back(i);
    }
  }
  f[0] = 1;
  for (int i = 1; i < 20005; ++i) {
    f[i] = (1ll * f[i - 1] * i) % md;
  }
  of[20004] = bin_pow(f[20004], md - 2);
  for (int i = 20003; i >= 0; --i) {
    of[i] = (1ll * of[i + 1] * (i + 1)) % md;
  }
  long long dd = d;
  for (int x : a) {
    while (dd % x == 0) dd /= x;
  }
  if (dd != 1) a.push_back(dd);
  int k = a.size();
  int tt;
  cin >> tt;
  while (tt--) {
    cin >> x >> y;
    if (x == y) {
      cout << 1 << '\n';
      continue;
    }
    vector<int> b(k, 0);
    for (int i = 0; i < k; ++i) {
      while (x % a[i] == 0) {
        x /= a[i];
        b[i]++;
      }
      while (y % a[i] == 0) {
        y /= a[i];
        b[i]--;
      }
    }
    int s = 0, sm = 0;
    for (int i = 0; i < k; ++i) {
      if (b[i] > 0)
        s += b[i];
      else
        sm -= b[i];
    }
    long long ans = (1ll * f[s] * f[sm]) % md;
    for (int i = 0; i < k; ++i) {
      ans = (1ll * ans * of[abs(b[i])]) % md;
    }
    cout << ans << '\n';
  }
}

### Prompt

Your challenge is to write 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;
const int N = 3e5 + 5;
const int md = 998244353;
long long d, x, y, f[20005], of[20005];
vector<long long> a;
bool ff(long long x) {
  long long fff = min((long long)sqrt(x) + 10, x - 1);
  for (long long i = 2; i <= fff; ++i) {
    if (x % i == 0) return 0;
  }
  return 1;
}
long long bin_pow(long long a, int b) {
  long long s = 1;
  while (b > 0) {
    if (b % 2 == 0) {
      a = (1ll * a * a) % md;
      b >>= 1;
    } else {
      s = (1ll * s * a) % md;
      b--;
    }
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> d;
  for (long long i = 2; i <= sqrt(d) + 10; ++i) {
    if (d % i == 0) {
      if (ff(i)) a.push_back(i);
    }
  }
  f[0] = 1;
  for (int i = 1; i < 20005; ++i) {
    f[i] = (1ll * f[i - 1] * i) % md;
  }
  of[20004] = bin_pow(f[20004], md - 2);
  for (int i = 20003; i >= 0; --i) {
    of[i] = (1ll * of[i + 1] * (i + 1)) % md;
  }
  long long dd = d;
  for (int x : a) {
    while (dd % x == 0) dd /= x;
  }
  if (dd != 1) a.push_back(dd);
  int k = a.size();
  int tt;
  cin >> tt;
  while (tt--) {
    cin >> x >> y;
    if (x == y) {
      cout << 1 << '\n';
      continue;
    }
    vector<int> b(k, 0);
    for (int i = 0; i < k; ++i) {
      while (x % a[i] == 0) {
        x /= a[i];
        b[i]++;
      }
      while (y % a[i] == 0) {
        y /= a[i];
        b[i]--;
      }
    }
    int s = 0, sm = 0;
    for (int i = 0; i < k; ++i) {
      if (b[i] > 0)
        s += b[i];
      else
        sm -= b[i];
    }
    long long ans = (1ll * f[s] * f[sm]) % md;
    for (int i = 0; i < k; ++i) {
      ans = (1ll * ans * of[abs(b[i])]) % md;
    }
    cout << ans << '\n';
  }
}

```