#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
const long long MOD = 998244353;
long long d, q;
vector<long long> prm;
long long dp_pow[1000];
long long mpow(long long a, long long n) {
  if (dp_pow[a] != -1) return dp_pow[a];
  if (n == 0) return 1;
  long long res = mpow(a, n / 2);
  res = res * res % MOD;
  if (n % 2) res = res * a % MOD;
  return dp_pow[a] = res;
}
map<long long, long long> mp;
long long dist(long long n) {
  if (mp.count(n)) return mp[n];
  long long tn = n;
  long long all_cnt = 0;
  vector<long long> v;
  for (long long p : prm) {
    long long cnt = 0;
    while (n % p == 0) {
      n /= p;
      cnt++;
    }
    all_cnt += cnt;
    if (cnt) v.push_back(cnt);
  }
  long long res = 1;
  for (long long i = 1; i <= all_cnt; i++) {
    res *= i;
    res %= MOD;
  }
  for (long long cnt : v) {
    for (long long i = 1; i <= cnt; i++) {
      res *= mpow(i, MOD - 2);
      res %= MOD;
    }
  }
  return mp[tn] = res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  memset(dp_pow, -1, sizeof dp_pow);
  cin >> d >> q;
  long long cd = d;
  for (long long i = 2; i * i <= d; i++) {
    long long cnt = 0;
    while (cd % i == 0) {
      cd /= i;
      cnt++;
    }
    if (cnt) prm.push_back(i);
  }
  if (cd != 1) prm.push_back(cd);
  while (q--) {
    long long u, v;
    cin >> u >> v;
    if (u == v) {
      cout << "1\n";
      continue;
    }
    long long g = gcd(u, v);
    long long ans = (dist(u / g) * dist(v / g)) % MOD;
    cout << ans << '\n';
  }
}

### Prompt

Please create a solution in Cpp to 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) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
const long long MOD = 998244353;
long long d, q;
vector<long long> prm;
long long dp_pow[1000];
long long mpow(long long a, long long n) {
  if (dp_pow[a] != -1) return dp_pow[a];
  if (n == 0) return 1;
  long long res = mpow(a, n / 2);
  res = res * res % MOD;
  if (n % 2) res = res * a % MOD;
  return dp_pow[a] = res;
}
map<long long, long long> mp;
long long dist(long long n) {
  if (mp.count(n)) return mp[n];
  long long tn = n;
  long long all_cnt = 0;
  vector<long long> v;
  for (long long p : prm) {
    long long cnt = 0;
    while (n % p == 0) {
      n /= p;
      cnt++;
    }
    all_cnt += cnt;
    if (cnt) v.push_back(cnt);
  }
  long long res = 1;
  for (long long i = 1; i <= all_cnt; i++) {
    res *= i;
    res %= MOD;
  }
  for (long long cnt : v) {
    for (long long i = 1; i <= cnt; i++) {
      res *= mpow(i, MOD - 2);
      res %= MOD;
    }
  }
  return mp[tn] = res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  memset(dp_pow, -1, sizeof dp_pow);
  cin >> d >> q;
  long long cd = d;
  for (long long i = 2; i * i <= d; i++) {
    long long cnt = 0;
    while (cd % i == 0) {
      cd /= i;
      cnt++;
    }
    if (cnt) prm.push_back(i);
  }
  if (cd != 1) prm.push_back(cd);
  while (q--) {
    long long u, v;
    cin >> u >> v;
    if (u == v) {
      cout << "1\n";
      continue;
    }
    long long g = gcd(u, v);
    long long ans = (dist(u / g) * dist(v / g)) % MOD;
    cout << ans << '\n';
  }
}

```