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 int mod = 998244353;
long long int binpow(long long int base, long long int exp) {
  base %= mod;
  long long int answer = 1;
  while (exp) {
    if (exp & 1) {
      answer *= base;
      answer %= mod;
    }
    base *= base;
    base %= mod;
    exp /= 2;
  }
  return answer;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int factorials[1000];
long long int inverseFactorials[1000];
long long int ways(vector<long long int>& freq) {
  long long int n = 0;
  for (auto& itr : freq) n += itr;
  long long int answer = factorials[n];
  for (auto& itr : freq) {
    answer = answer * inverseFactorials[itr];
    answer %= mod;
  }
  return answer;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  factorials[0] = 1;
  inverseFactorials[0] = 1;
  for (long long int i = 1; i < 1000; i++) {
    factorials[i] = factorials[i - 1] * i;
    factorials[i] %= mod;
    inverseFactorials[i] = binpow(factorials[i], mod - 2);
  }
  long long int d;
  cin >> d;
  set<long long int> primes;
  long long int n = d;
  while (n % 2 == 0) {
    primes.insert(2);
    n /= 2;
  }
  for (long long int i = 3; i * i <= n; i += 2) {
    while (n % i == 0) {
      primes.insert(i);
      n /= i;
    }
  }
  if (n > 2) primes.insert(n);
  vector<long long int> primeFactors(primes.begin(), primes.end());
  n = primeFactors.size();
  long long int q;
  cin >> q;
  while (q--) {
    long long int a, b;
    cin >> a >> b;
    vector<long long int> hash1(n, 0);
    vector<long long int> hash2(n, 0);
    long long int c = gcd(a, b);
    a /= c;
    b /= c;
    for (long long int i = 0; i < n; i++) {
      while (a % primeFactors[i] == 0) {
        a /= primeFactors[i];
        hash1[i]++;
      }
    }
    for (long long int i = 0; i < n; i++) {
      while (b % primeFactors[i] == 0) {
        b /= primeFactors[i];
        hash2[i]++;
      }
    }
    long long int waysA = ways(hash1), waysB = ways(hash2);
    long long int answer = waysA * waysB;
    answer %= mod;
    cout << answer << '\n';
  }
  return 0;
}