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>
long long int gcd(long long int a, long long int b) {
  return a ? gcd(b % a, a) : b;
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  long long int s = a;
  long long int t = b;
  while (t) {
    if (t % 2 == 1) ans *= s, ans %= 998244353;
    s *= s, s %= 998244353;
    t /= 2;
  }
  return ans;
}
long long int inv(long long int k) { return power(k, 998244353 - 2); }
long long int fact[110];
long long int invFact[110];
std::vector<long long int> prime;
int check[110];
std::vector<int> V;
long long int func(long long int k) {
  V.clear();
  for (int i = 0; i < prime.size(); i++) {
    if (k % prime[i] != 0) continue;
    int count = 0;
    while (k % prime[i] == 0) count++, k /= prime[i];
    V.push_back(count);
  }
  int sum = 0;
  for (int i = 0; i < V.size(); i++) sum += V[i];
  long long int ans = fact[sum];
  for (int i = 0; i < V.size(); i++) ans *= invFact[V[i]], ans %= 998244353;
  return ans;
}
std::vector<long long int> P;
int main() {
  fact[0] = 1;
  for (int i = 1; i <= 100; i++)
    fact[i] = fact[i - 1] * i, fact[i] %= 998244353;
  for (int i = 0; i <= 100; i++) invFact[i] = inv(fact[i]);
  long long int a;
  int b;
  scanf("%lld%d", &a, &b);
  long long int a2 = a;
  for (long long int i = 2; i * i <= a; i++) {
    if (a2 % i == 0) {
      prime.push_back(i);
      while (a2 % i == 0) a2 /= i;
    }
  }
  if (a2 > 1) prime.push_back(a2);
  for (int i = 1; i <= b; i++) {
    long long int c, d;
    scanf("%lld%lld", &c, &d);
    long long int e = gcd(c, d);
    long long int s1 = func(c / e);
    long long int s2 = func(d / e);
    printf("%lld\n", (s1 * s2) % 998244353);
  }
}