#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long fastxp(long long b, long long e) {
  if (e == 1)
    return b;
  else {
    long long ans = fastxp(b, e / 2);
    ans = (ans * ans) % 998244353;
    if (e % 2 == 1) ans = (ans * b) % 998244353;
    return ans;
  }
}
long long fac[100010], ifac[100010];
vector<long long> p;
map<long long, long long> m;
long long d, q;
void build() {
  fac[0] = 1;
  for (long long i = 1; i <= 100000; i++) {
    fac[i] = (fac[i - 1] * i) % 998244353;
    ifac[i] = fastxp(fac[i], 998244353 - 2);
  }
  long long i = 2;
  while (i * i <= d) {
    if (d % i == 0) {
      p.push_back(i);
      while (i * i <= d && d % i == 0) d /= i;
    }
    i++;
  }
  if (d > 1) p.push_back(d);
}
long long ans(long long a) {
  if (m[a] != 0) {
    return m[a];
  }
  long long ret = 0, A = a;
  vector<long long> v;
  for (long long x : p) {
    if (a % x == 0) {
      long long aux = 0;
      while (a % x == 0) {
        a /= x;
        aux++;
      }
      ret += aux;
      v.push_back(aux);
    }
  }
  ret = fac[ret];
  for (long long x : v) {
    ret = (ret * ifac[x]) % 998244353;
  }
  m[A] = ret;
  return ret;
}
int main() {
  scanf("%lld %lld", &d, &q);
  build();
  while (q--) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    long long g = gcd(a, b);
    printf("%lld\n", (ans(a / g) * ans(b / g)) % 998244353);
  }
}

### 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;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long fastxp(long long b, long long e) {
  if (e == 1)
    return b;
  else {
    long long ans = fastxp(b, e / 2);
    ans = (ans * ans) % 998244353;
    if (e % 2 == 1) ans = (ans * b) % 998244353;
    return ans;
  }
}
long long fac[100010], ifac[100010];
vector<long long> p;
map<long long, long long> m;
long long d, q;
void build() {
  fac[0] = 1;
  for (long long i = 1; i <= 100000; i++) {
    fac[i] = (fac[i - 1] * i) % 998244353;
    ifac[i] = fastxp(fac[i], 998244353 - 2);
  }
  long long i = 2;
  while (i * i <= d) {
    if (d % i == 0) {
      p.push_back(i);
      while (i * i <= d && d % i == 0) d /= i;
    }
    i++;
  }
  if (d > 1) p.push_back(d);
}
long long ans(long long a) {
  if (m[a] != 0) {
    return m[a];
  }
  long long ret = 0, A = a;
  vector<long long> v;
  for (long long x : p) {
    if (a % x == 0) {
      long long aux = 0;
      while (a % x == 0) {
        a /= x;
        aux++;
      }
      ret += aux;
      v.push_back(aux);
    }
  }
  ret = fac[ret];
  for (long long x : v) {
    ret = (ret * ifac[x]) % 998244353;
  }
  m[A] = ret;
  return ret;
}
int main() {
  scanf("%lld %lld", &d, &q);
  build();
  while (q--) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    long long g = gcd(a, b);
    printf("%lld\n", (ans(a / g) * ans(b / g)) % 998244353);
  }
}

```