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;
template <class T>
inline void read(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return;
}
inline void out(long long x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
const int mod = 998244353;
const int maxn = 1e5 + 10;
map<long long, int> mp;
long long fac[maxn] = {0};
long long quickpow(long long a, long long b) {
  long long ans = 1, base = a;
  while (b) {
    if (b & 1) ans = ans * base % mod;
    base = base * base % mod;
    b >>= 1;
  }
  return ans;
}
long long C(int m, int n) {
  return fac[n] * quickpow(fac[m] * fac[n - m] % mod, mod - 2) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int cal(long long n) {
  vector<int> v;
  for (long long i = 2; i <= sqrt(n); i++) {
    int cnt = 0;
    while (n % i == 0) cnt++, n /= i;
    if (cnt) v.push_back(cnt);
  }
  if (n > 1) v.push_back(1);
  long long ans = 1;
  int sum = 0;
  for (int i = 0; i < v.size(); i++) {
    int x = v[i];
    sum += x;
    ans = ans * C(x, sum) % mod;
  }
  return ans;
}
int main() {
  long long d, q, u, v;
  read(d);
  fac[1] = 1, fac[0] = 1;
  for (int i = 2; i < maxn; i++) {
    fac[i] = fac[i - 1] * i % mod;
  }
  read(q);
  while (q--) {
    read(u), read(v);
    long long gc = gcd(u, v);
    if (mp[u / gc] == 0) mp[u / gc] = cal(u / gc);
    if (mp[v / gc] == 0) mp[v / gc] = cal(v / gc);
    printf("%lld\n", (1ll * mp[u / gc] * mp[v / gc] + mod) % mod);
  }
  return 0;
}