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;
int q;
long long d, u, v;
long long pri[1000010], pricnt;
long long cnt, cntt, tot, fac[1000010];
long long fpow(long long a, long long b, long long m) {
  long long ans = 1;
  a %= m;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans % m;
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  scanf("%lld%d", &d, &q);
  fac[0] = 1;
  for (int i = 1; i <= 10010; i++) fac[i] = fac[i - 1] * i % mod;
  for (long long i = 2; i * i <= d; i++) {
    if (d % i == 0) pri[++pricnt] = i;
    while (d % i == 0) d = d / i;
  }
  if (d > 1) pri[++pricnt] = d;
  while (q--) {
    scanf("%lld%lld", &u, &v);
    long long nb = gcd(u, v);
    long long p1 = u / nb, p2 = v / nb;
    long long fucksb = 1;
    cnt = 0;
    tot = 0;
    for (int i = 1; i <= pricnt; i++) {
      if (p1 % pri[i] != 0) continue;
      tot++;
      while (p1 % pri[i] == 0) {
        p1 /= pri[i];
        cntt++;
      }
      fucksb = fucksb * fac[cntt] % mod;
      cnt += cntt;
      cntt = 0;
    }
    if (p1 > 1) {
      tot++;
      cnt++;
      p1 = 1;
    }
    long long ans1 = fac[cnt];
    ans1 = ans1 * fpow(fucksb, mod - 2, mod) % mod;
    cnt = 0;
    tot = 0;
    fucksb = 1;
    for (int i = 1; i <= pricnt; i++) {
      if (p2 % pri[i] != 0) continue;
      tot++;
      while (p2 % pri[i] == 0) {
        p2 /= pri[i];
        cntt++;
      }
      fucksb = fucksb * fac[cntt] % mod;
      cnt += cntt;
      cntt = 0;
    }
    if (p2 > 1) {
      tot++;
      cnt++;
      p2 = 1;
    }
    long long ans2 = fac[cnt];
    ans2 = ans2 * fpow(fucksb, mod - 2, mod) % mod;
    printf("%lld\n", ans1 * ans2 % mod);
  }
}