#include <bits/stdc++.h>
using namespace std;
void fastIO() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
template <typename T, typename U>
inline void mnze(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void mxze(T &x, U y) {
  if (x < y) x = y;
}
void _scan(int &x) { scanf("%d", &x); }
void _scan(long long &x) { scanf("%lld", &x); }
void _scan(double &x) { scanf("%lf", &x); }
void _scan(char &x) { scanf(" %c", &x); }
void _scan(char *x) { scanf("%s", x); }
template <typename T>
void _scan(vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++) _scan(v[i]);
}
void scan() {}
template <typename T, typename... U>
void scan(T &head, U &...tail) {
  _scan(head);
  scan(tail...);
}
template <typename T>
void _dbg(const char *sdbg, T h) {
  cerr << sdbg << "=" << h << "\n";
}
template <typename T, typename... U>
void _dbg(const char *sdbg, T h, U... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, t...);
}
long long MOD = 998244353;
inline long long Msum(long long x) { return x; }
template <typename... Rest>
inline long long Msum(long long x, Rest... rest) {
  return (x + Msum(rest...)) % MOD;
}
inline long long Mprod(long long x) { return x; }
template <typename... Rest>
inline long long Mprod(long long x, Rest... rest) {
  return x * Mprod(rest...) % MOD;
}
inline long long Mnorm(long long x) { return (x % MOD + MOD) % MOD; }
inline long long sq(long long x) { return x * x % MOD; }
inline long long pot(long long b, long long e) {
  if (!e) return 1;
  return sq(pot(b, e >> 1)) * (e & 1 ? b : 1) % MOD;
}
inline long long inv(long long x) { return pot(x, MOD - 2); }
vector<long long> p;
vector<long long> geta(long long x) {
  vector<long long> r(p.size());
  for (int i = 0; i < (int)p.size(); i++) {
    while (x % p[i] == 0) {
      x /= p[i];
      r[i]++;
    }
  }
  return (r);
}
long long fact[1000];
int main() {
  long long D;
  scan(D);
  long long aux = D;
  for (long long d = 2; d * d <= aux; d++) {
    if (aux % d == 0) {
      p.push_back(d);
      while (aux % d == 0) {
        aux /= d;
      }
    }
  }
  if (aux > 1) p.push_back(aux);
  fact[0] = 1;
  for (int i = 1; i < 1000; i++) fact[i] = fact[i - 1] * i % MOD;
  int q;
  scan(q);
  while (q--) {
    long long x, y;
    scan(x, y);
    vector<long long> a = geta(x);
    vector<long long> b = geta(y);
    long long p = 0, q = 1;
    long long p2 = 0, q2 = 1;
    for (int i = 0; i < (int)a.size(); i++) {
      int d = a[i] - b[i];
      if (d >= 0) {
        q = q * fact[d] % MOD;
        p = p + d;
      } else {
        d = -d;
        q2 = q2 * fact[d] % MOD;
        p2 = p2 + d;
      }
    }
    p = fact[p];
    p2 = fact[p2];
    long long ans = (p * inv(q) % MOD) * (p2 * inv(q2) % MOD);
    printf("%lld\n", ans % MOD);
  }
}

### Prompt

Create a solution in Cpp for 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;
void fastIO() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
template <typename T, typename U>
inline void mnze(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void mxze(T &x, U y) {
  if (x < y) x = y;
}
void _scan(int &x) { scanf("%d", &x); }
void _scan(long long &x) { scanf("%lld", &x); }
void _scan(double &x) { scanf("%lf", &x); }
void _scan(char &x) { scanf(" %c", &x); }
void _scan(char *x) { scanf("%s", x); }
template <typename T>
void _scan(vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++) _scan(v[i]);
}
void scan() {}
template <typename T, typename... U>
void scan(T &head, U &...tail) {
  _scan(head);
  scan(tail...);
}
template <typename T>
void _dbg(const char *sdbg, T h) {
  cerr << sdbg << "=" << h << "\n";
}
template <typename T, typename... U>
void _dbg(const char *sdbg, T h, U... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, t...);
}
long long MOD = 998244353;
inline long long Msum(long long x) { return x; }
template <typename... Rest>
inline long long Msum(long long x, Rest... rest) {
  return (x + Msum(rest...)) % MOD;
}
inline long long Mprod(long long x) { return x; }
template <typename... Rest>
inline long long Mprod(long long x, Rest... rest) {
  return x * Mprod(rest...) % MOD;
}
inline long long Mnorm(long long x) { return (x % MOD + MOD) % MOD; }
inline long long sq(long long x) { return x * x % MOD; }
inline long long pot(long long b, long long e) {
  if (!e) return 1;
  return sq(pot(b, e >> 1)) * (e & 1 ? b : 1) % MOD;
}
inline long long inv(long long x) { return pot(x, MOD - 2); }
vector<long long> p;
vector<long long> geta(long long x) {
  vector<long long> r(p.size());
  for (int i = 0; i < (int)p.size(); i++) {
    while (x % p[i] == 0) {
      x /= p[i];
      r[i]++;
    }
  }
  return (r);
}
long long fact[1000];
int main() {
  long long D;
  scan(D);
  long long aux = D;
  for (long long d = 2; d * d <= aux; d++) {
    if (aux % d == 0) {
      p.push_back(d);
      while (aux % d == 0) {
        aux /= d;
      }
    }
  }
  if (aux > 1) p.push_back(aux);
  fact[0] = 1;
  for (int i = 1; i < 1000; i++) fact[i] = fact[i - 1] * i % MOD;
  int q;
  scan(q);
  while (q--) {
    long long x, y;
    scan(x, y);
    vector<long long> a = geta(x);
    vector<long long> b = geta(y);
    long long p = 0, q = 1;
    long long p2 = 0, q2 = 1;
    for (int i = 0; i < (int)a.size(); i++) {
      int d = a[i] - b[i];
      if (d >= 0) {
        q = q * fact[d] % MOD;
        p = p + d;
      } else {
        d = -d;
        q2 = q2 * fact[d] % MOD;
        p2 = p2 + d;
      }
    }
    p = fact[p];
    p2 = fact[p2];
    long long ans = (p * inv(q) % MOD) * (p2 * inv(q2) % MOD);
    printf("%lld\n", ans % MOD);
  }
}

```