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;
vector<long long> v;
long long a[100], b[100];
long long c[100000][100];
long long f[100000], g[100000];
const int mod = 998244353;
long long qpow(long long a, long long n) {
  long long re = 1;
  while (n > 0) {
    if (n & 1) {
      re *= a;
      re %= mod;
    }
    a *= a;
    a %= mod;
    n >>= 1;
  }
  return re;
}
int main() {
  f[0] = g[0] = 1;
  for (long long i = 1; i < 100000; i++) {
    f[i] = f[i - 1] * i;
    f[i] %= mod;
    g[i] = qpow(f[i], mod - 2);
  }
  long long d;
  scanf("%lld", &d);
  int q;
  scanf("%d", &q);
  if (d == 1) {
    while (q--) {
      printf("1\n");
    }
    return 0;
  }
  long long temp = d;
  int now = 0;
  for (int i = 2; i <= sqrt(d); i++) {
    if (temp % i == 0) {
      a[++now] = i;
      while (temp % i == 0) {
        b[now]++;
        temp /= i;
      }
    }
  }
  if (temp > 1) {
    a[++now] = temp;
    b[now]++;
  }
  for (int i = 1; i <= sqrt(d); i++) {
    if (d % i == 0) {
      v.push_back(i);
      v.push_back(d / i);
    }
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size() / 2; i++) {
    long long temp = v[i];
    for (int j = 1; j <= now; j++) {
      while (temp % a[j] == 0) {
        temp /= a[j];
        c[i][j]++;
      }
      c[v.size() - 1 - i][j] = b[j] - c[i][j];
    }
  }
  while (q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    int pos1 = lower_bound(v.begin(), v.end(), x) - v.begin();
    int pos2 = lower_bound(v.begin(), v.end(), y) - v.begin();
    long long ans = 1;
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= now; i++) {
      if (c[pos1][i] == c[pos2][i]) {
        continue;
      } else if (c[pos1][i] > c[pos2][i]) {
        cnt1 += abs(c[pos1][i] - c[pos2][i]);
      } else if (c[pos2][i] > c[pos1][i]) {
        cnt2 += abs(c[pos1][i] - c[pos2][i]);
      }
      ans *= g[abs(c[pos1][i] - c[pos2][i])];
      ans %= mod;
    }
    ans *= f[cnt1];
    ans %= mod;
    ans *= f[cnt2];
    ans %= mod;
    printf("%lld\n", ans);
  }
}