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;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const int MOD = 998244353;
int sum(int a, int b) {
  int out = a + b;
  if (out >= MOD) out -= MOD;
  return out;
}
int prod(int a, int b) { return ((long long)a * b) % MOD; }
int bpow(int a, int b) {
  int out = 1;
  while (b > 0) {
    if (b & 1) out = prod(out, a);
    a = prod(a, a);
    b >>= 1;
  }
  return out;
}
int inv(int a) { return bpow(a, MOD - 2); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long d;
  cin >> d;
  vector<long long> primes;
  long long n = d;
  for (long long p = 2; (long long)p * p <= n; ++p) {
    if (n % p == 0) {
      int pwr = 0;
      while (n % p == 0) {
        n /= p;
        ++pwr;
      }
      primes.push_back(p);
    }
  }
  if (n > 1) {
    primes.push_back(n);
  }
  int fact[60], inv_fact[60];
  fact[0] = 1;
  for (int i = 1; i < 60; ++i) fact[i] = prod(fact[i - 1], i);
  for (int i = 0; i < 60; ++i) inv_fact[i] = inv(fact[i]);
  int q;
  cin >> q;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    int wu = 1, wv = 1, su = 0, sv = 0;
    for (auto p : primes) {
      int pu = 0, pv = 0;
      while (u % p == 0) {
        u /= p;
        ++pu;
      }
      while (v % p == 0) {
        v /= p;
        ++pv;
      }
      if (pu > pv) {
        su += pu - pv;
        wu = prod(wu, inv_fact[pu - pv]);
      }
      if (pv > pu) {
        sv += pv - pu;
        wv = prod(wv, inv_fact[pv - pu]);
      }
    }
    wu = prod(wu, fact[su]);
    wv = prod(wv, fact[sv]);
    cout << prod(wu, wv) << '\n';
  }
}