#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e5 + 10, mod = 998244353, lim = 2e5;
int m, cnt, tot;
int t[maxN + 1][22], s[22], tmp[22];
long long p[22], fac[maxN + 1], inv[maxN + 1];
long long n;
map<long long, int> f[maxN + 1], id;
inline int mpow(int a, int x) {
  int ans = 1;
  while (x) {
    if (x & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    x >>= 1;
  }
  return ans;
}
inline void dfs(int x, long long sum) {
  if (x > tot) {
    id[sum] = ++cnt;
    for (int i = 1; i <= tot; i++) t[cnt][i] = tmp[i];
    return;
  }
  long long r = 1;
  for (int i = 0; i <= s[x]; i++, r *= (long long)p[x])
    tmp[x] = i, dfs(x + 1, sum * r);
}
int main() {
  scanf("%lld", &n);
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) {
      p[++tot] = i;
      s[tot] = 0;
      while (n % i == 0) s[tot]++, n /= i;
    }
  if (n > 1) p[++tot] = n, s[tot] = 1;
  dfs(1, 1);
  fac[0] = 1;
  for (int i = 1; i <= lim; i++) fac[i] = fac[i - 1] * i % mod;
  inv[lim] = mpow(fac[lim], mod - 2);
  for (int i = lim - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  scanf("%d", &m);
  while (m--) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    long long ans = 1, s1 = 0, s2 = 0;
    x = id[x], y = id[y];
    for (int i = 1; i <= tot; i++) {
      int s = t[x][i] - t[y][i];
      if (!s) continue;
      if (s > 0)
        ans = ans * inv[s] % mod, s1 += s;
      else
        ans = ans * inv[-s] % mod, s2 -= s;
    }
    printf("%lld\n", ans * fac[s1] % mod * fac[s2] % mod);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 int maxN = 2e5 + 10, mod = 998244353, lim = 2e5;
int m, cnt, tot;
int t[maxN + 1][22], s[22], tmp[22];
long long p[22], fac[maxN + 1], inv[maxN + 1];
long long n;
map<long long, int> f[maxN + 1], id;
inline int mpow(int a, int x) {
  int ans = 1;
  while (x) {
    if (x & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    x >>= 1;
  }
  return ans;
}
inline void dfs(int x, long long sum) {
  if (x > tot) {
    id[sum] = ++cnt;
    for (int i = 1; i <= tot; i++) t[cnt][i] = tmp[i];
    return;
  }
  long long r = 1;
  for (int i = 0; i <= s[x]; i++, r *= (long long)p[x])
    tmp[x] = i, dfs(x + 1, sum * r);
}
int main() {
  scanf("%lld", &n);
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) {
      p[++tot] = i;
      s[tot] = 0;
      while (n % i == 0) s[tot]++, n /= i;
    }
  if (n > 1) p[++tot] = n, s[tot] = 1;
  dfs(1, 1);
  fac[0] = 1;
  for (int i = 1; i <= lim; i++) fac[i] = fac[i - 1] * i % mod;
  inv[lim] = mpow(fac[lim], mod - 2);
  for (int i = lim - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  scanf("%d", &m);
  while (m--) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    long long ans = 1, s1 = 0, s2 = 0;
    x = id[x], y = id[y];
    for (int i = 1; i <= tot; i++) {
      int s = t[x][i] - t[y][i];
      if (!s) continue;
      if (s > 0)
        ans = ans * inv[s] % mod, s1 += s;
      else
        ans = ans * inv[-s] % mod, s2 -= s;
    }
    printf("%lld\n", ans * fac[s1] % mod * fac[s2] % mod);
  }
  return 0;
}

```