#include <bits/stdc++.h>
using LL = long long;
using namespace std;
constexpr LL MOD = 998244353;
int binom[51][51];
LL gcd(LL a, LL b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
vector<int> C;
LL cnt(LL a, const vector<LL> &P) {
  C.clear();
  int s = 0;
  for (LL p : P) {
    int c = 0;
    while (a % p == 0) {
      a /= p;
      ++c;
    }
    C.push_back(c);
    s += c;
  }
  LL ans = 1;
  for (int c : C) {
    ans = ans * binom[s][c] % MOD;
    s -= c;
  }
  return ans;
}
int main() {
  for (int i = 0; i <= 50; ++i) {
    for (int j = 0; j <= i; ++j) {
      if (j == 0 || j == i) {
        binom[i][j] = 1;
      } else {
        binom[i][j] = ((LL)binom[i - 1][j] + binom[i - 1][j - 1]) % MOD;
      }
    }
  }
  LL D;
  int Q;
  cin >> D >> Q;
  vector<LL> P;
  LL p = 2;
  while (D > 1) {
    while (D % p != 0 && (LL)p * p < D) {
      ++p;
    }
    if (D % p != 0) {
      p = D;
    }
    P.push_back(p);
    while (D % p == 0) {
      D /= p;
    }
  }
  for (int q = 0; q < Q; ++q) {
    LL u, v;
    cin >> u >> v;
    LL g = gcd(u, v);
    LL a = u / g, b = v / g;
    cout << cnt(a, P) * cnt(b, P) % MOD << '\n';
  }
  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 LL = long long;
using namespace std;
constexpr LL MOD = 998244353;
int binom[51][51];
LL gcd(LL a, LL b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
vector<int> C;
LL cnt(LL a, const vector<LL> &P) {
  C.clear();
  int s = 0;
  for (LL p : P) {
    int c = 0;
    while (a % p == 0) {
      a /= p;
      ++c;
    }
    C.push_back(c);
    s += c;
  }
  LL ans = 1;
  for (int c : C) {
    ans = ans * binom[s][c] % MOD;
    s -= c;
  }
  return ans;
}
int main() {
  for (int i = 0; i <= 50; ++i) {
    for (int j = 0; j <= i; ++j) {
      if (j == 0 || j == i) {
        binom[i][j] = 1;
      } else {
        binom[i][j] = ((LL)binom[i - 1][j] + binom[i - 1][j - 1]) % MOD;
      }
    }
  }
  LL D;
  int Q;
  cin >> D >> Q;
  vector<LL> P;
  LL p = 2;
  while (D > 1) {
    while (D % p != 0 && (LL)p * p < D) {
      ++p;
    }
    if (D % p != 0) {
      p = D;
    }
    P.push_back(p);
    while (D % p == 0) {
      D /= p;
    }
  }
  for (int q = 0; q < Q; ++q) {
    LL u, v;
    cin >> u >> v;
    LL g = gcd(u, v);
    LL a = u / g, b = v / g;
    cout << cnt(a, P) * cnt(b, P) % MOD << '\n';
  }
  return 0;
}

```