#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long qpow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod, y >>= 1;
  }
  return ans;
}
int T, cnt, now[110];
long long u, v, d, p[110], frac[110], inv[110];
int C(int x, int y) { return frac[x] * inv[y] % mod * inv[x - y] % mod; }
long long solve(long long x) {
  int res = 0, tot = 0;
  long long ans = 1;
  for (int i = 1; i <= cnt; i++) {
    if (x % p[i] == 0) now[++tot] = 0;
    while (x % p[i] == 0) x /= p[i], now[tot]++, res++;
  }
  for (int i = 1; i <= tot; i++) {
    ans = ans * C(res, now[i]) % mod;
    res -= now[i];
  }
  return ans;
}
int main() {
  frac[0] = inv[0] = 1;
  for (int i = 1; i <= 100; i++) frac[i] = frac[i - 1] * i % mod;
  for (int i = 1; i <= 100; i++) inv[i] = qpow(frac[i], mod - 2);
  scanf("%lld%d", &d, &T);
  for (long long i = 2; i * i <= d; i++) {
    if (d % i == 0) p[++cnt] = i;
    while (d % i == 0) d /= i;
  }
  if (d > 1) p[++cnt] = d;
  while (T--) {
    scanf("%lld%lld", &u, &v);
    long long g = gcd(u, v);
    u /= g, v /= g;
    printf("%lld\n", solve(u) * solve(v) % mod);
  }
}

### Prompt

In Cpp, your task is to solve the following 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). 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long qpow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod, y >>= 1;
  }
  return ans;
}
int T, cnt, now[110];
long long u, v, d, p[110], frac[110], inv[110];
int C(int x, int y) { return frac[x] * inv[y] % mod * inv[x - y] % mod; }
long long solve(long long x) {
  int res = 0, tot = 0;
  long long ans = 1;
  for (int i = 1; i <= cnt; i++) {
    if (x % p[i] == 0) now[++tot] = 0;
    while (x % p[i] == 0) x /= p[i], now[tot]++, res++;
  }
  for (int i = 1; i <= tot; i++) {
    ans = ans * C(res, now[i]) % mod;
    res -= now[i];
  }
  return ans;
}
int main() {
  frac[0] = inv[0] = 1;
  for (int i = 1; i <= 100; i++) frac[i] = frac[i - 1] * i % mod;
  for (int i = 1; i <= 100; i++) inv[i] = qpow(frac[i], mod - 2);
  scanf("%lld%d", &d, &T);
  for (long long i = 2; i * i <= d; i++) {
    if (d % i == 0) p[++cnt] = i;
    while (d % i == 0) d /= i;
  }
  if (d > 1) p[++cnt] = d;
  while (T--) {
    scanf("%lld%lld", &u, &v);
    long long g = gcd(u, v);
    u /= g, v /= g;
    printf("%lld\n", solve(u) * solve(v) % mod);
  }
}

```