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). 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T> v) {
  os << "(" << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
string num2bit(long long num, long long len) {
  string bit = "";
  for (int i = 0; i < (int)len; ++i) {
    bit += char('0' + (num >> i & 1));
  }
  return bit;
}
map<long long, long long> mp;
void prime_factorization(long long x) {
  bool flag = true;
  while (flag) {
    long long sqx = sqrt(x) + 1;
    for (long long j = 2; j < sqx + 1; j++) {
      if (x % j == 0) {
        mp[j]++;
        x /= j;
        break;
      }
      if (j == sqx) flag = false;
    }
  }
  if (x != 1) {
    mp[x]++;
  }
}
const long long N_MAX = 100005;
long long inv[N_MAX], fac[N_MAX], finv[N_MAX];
void make() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < N_MAX; i++) {
    inv[i] = mod - inv[mod % i] * (mod / i) % mod;
    fac[i] = fac[i - 1] * (long long)i % mod;
    finv[i] = finv[i - 1] * inv[i] % mod;
  }
}
long long combination(long long C, long long D) {
  if (C < D || C < 0) return 0;
  return fac[C] * (finv[D] * finv[C - D] % mod) % mod;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  make();
  long long D;
  cin >> D;
  prime_factorization(D);
  vector<long long> yaku;
  for (auto x : mp) yaku.push_back(x.first);
  long long n = yaku.size();
  long long q;
  cin >> q;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    vector<long long> yu(n, 0), yv(n, 0), ym(n, 0);
    for (int i = 0; i < (int)n; ++i) {
      while (u % yaku[i] == 0) {
        u /= yaku[i];
        yu[i]++;
      }
      while (v % yaku[i] == 0) {
        v /= yaku[i];
        yv[i]++;
      }
      ym[i] = min(yu[i], yv[i]);
      yu[i] -= ym[i];
      yv[i] -= ym[i];
    }
    long long res = 1;
    long long tmpu = 0, tmpv = 0;
    for (int i = 0; i < (int)n; ++i) {
      tmpu += yu[i];
      res *= combination(tmpu, yu[i]);
      res %= mod;
      tmpv += yv[i];
      res *= combination(tmpv, yv[i]);
      res %= mod;
    }
    cout << res << "\n";
  }
  return 0;
}