#include <bits/stdc++.h>
using namespace std;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
static const ULL M = 998244353;
struct Node {
  ULL v;
  ULL d;
  ULL m;
};
struct Comb {
  ULL F[51];
  ULL I[51];
  ULL iF[51];
  Comb() {
    F[0] = 1;
    for (UL i = 1; i <= 50; i++) F[i] = F[i - 1] * i % M;
    I[1] = 1;
    for (UL i = 2; i <= 50; i++) I[i] = M - M / i * I[M % i] % M;
    iF[0] = 1;
    for (UL i = 1; i <= 50; i++) iF[i] = iF[i - 1] * I[i] % M;
  }
  ULL operator()(UL n, UL r) { return F[n] * iF[r] % M * iF[n - r] % M; }
};
ULL GCD(ULL a, ULL b) { return b ? GCD(b, a % b) : a; }
struct Problem {
  void Solve() {
    ULL D;
    scanf("%lld", &D);
    vector<pair<ULL, UL>> F;
    for (ULL i = 2; i * i <= D; i++) {
      if (D % i) continue;
      F.push_back({i, 0});
      while (D % i == 0) {
        D /= i;
        F.back().second++;
      }
    }
    if (D != 1) F.push_back({D, 1});
    Comb C;
    vector<Node> Dv = {{1, 0, 1}};
    for (auto& f : F) {
      UL S = Dv.size();
      for (UL i = 0; i < (f.second); i++) {
        for (UL j = 0; j < (S); j++) {
          Node tmp = Dv[j + i * S];
          tmp.v *= f.first;
          tmp.d = Dv[j].d + (i + 1);
          tmp.m = Dv[j].m * C(tmp.d, Dv[j].d) % M;
          Dv.push_back(tmp);
        }
      }
    }
    map<ULL, UL> G;
    for (UL i = 0; i < Dv.size(); i++) G[Dv[i].v] = i;
    UL Q;
    scanf("%d", &Q);
    while (Q--) {
      ULL u, v;
      scanf("%lld%lld", &u, &v);
      ULL g = GCD(u, v);
      u /= g;
      v /= g;
      ULL ans = Dv[G[u]].m * Dv[G[v]].m % M;
      printf("%lld\n", ans);
    }
  }
};
int main() {
  unique_ptr<Problem> p(new Problem());
  p->Solve();
  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;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
static const ULL M = 998244353;
struct Node {
  ULL v;
  ULL d;
  ULL m;
};
struct Comb {
  ULL F[51];
  ULL I[51];
  ULL iF[51];
  Comb() {
    F[0] = 1;
    for (UL i = 1; i <= 50; i++) F[i] = F[i - 1] * i % M;
    I[1] = 1;
    for (UL i = 2; i <= 50; i++) I[i] = M - M / i * I[M % i] % M;
    iF[0] = 1;
    for (UL i = 1; i <= 50; i++) iF[i] = iF[i - 1] * I[i] % M;
  }
  ULL operator()(UL n, UL r) { return F[n] * iF[r] % M * iF[n - r] % M; }
};
ULL GCD(ULL a, ULL b) { return b ? GCD(b, a % b) : a; }
struct Problem {
  void Solve() {
    ULL D;
    scanf("%lld", &D);
    vector<pair<ULL, UL>> F;
    for (ULL i = 2; i * i <= D; i++) {
      if (D % i) continue;
      F.push_back({i, 0});
      while (D % i == 0) {
        D /= i;
        F.back().second++;
      }
    }
    if (D != 1) F.push_back({D, 1});
    Comb C;
    vector<Node> Dv = {{1, 0, 1}};
    for (auto& f : F) {
      UL S = Dv.size();
      for (UL i = 0; i < (f.second); i++) {
        for (UL j = 0; j < (S); j++) {
          Node tmp = Dv[j + i * S];
          tmp.v *= f.first;
          tmp.d = Dv[j].d + (i + 1);
          tmp.m = Dv[j].m * C(tmp.d, Dv[j].d) % M;
          Dv.push_back(tmp);
        }
      }
    }
    map<ULL, UL> G;
    for (UL i = 0; i < Dv.size(); i++) G[Dv[i].v] = i;
    UL Q;
    scanf("%d", &Q);
    while (Q--) {
      ULL u, v;
      scanf("%lld%lld", &u, &v);
      ULL g = GCD(u, v);
      u /= g;
      v /= g;
      ULL ans = Dv[G[u]].m * Dv[G[v]].m % M;
      printf("%lld\n", ans);
    }
  }
};
int main() {
  unique_ptr<Problem> p(new Problem());
  p->Solve();
  return 0;
}

```