#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int N = 500031;
long long D;
vector<long long> primes;
long long gcd(long long a, long long b) {
  while (a && b) a > b ? a %= b : b %= a;
  return a + b;
}
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2) return a * pw(a, b - 1) % 998244353;
  return pw(a * a % 998244353, b / 2);
}
long long inv(long long x) { return pw(x, 998244353 - 2); }
int tests;
long long A, B;
long long ans;
long long fact[N], invf[N];
int count_deg(long long a, long long b) {
  int ret = 0;
  while (a % b == 0) {
    ret++;
    a /= b;
  }
  return ret;
}
long long C(long long n, long long m) {
  long long res = fact[n] * invf[m];
  res %= 998244353;
  res *= invf[n - m];
  return res % 998244353;
}
long long solver(long long a, long long b) {
  long long res = 1;
  long long ttl = 0;
  for (int i = 0; i < primes.size(); i++) {
    int dif = abs(count_deg(a, primes[i]) - count_deg(b, primes[i]));
    res = res * C(ttl + dif, ttl);
    res %= 998244353;
    ttl += dif;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  fact[0] = 1;
  for (int i = 1; i < 100; i++) {
    fact[i] = fact[i - 1] * i % 998244353;
  }
  for (int i = 0; i < 100; i++) {
    invf[i] = inv(fact[i]);
  }
  cin >> D;
  if (D % 2 == 0) {
    primes.push_back(2);
    while (D % 2 == 0) {
      D /= 2;
    }
  }
  for (int i = 3; 1ll * i * i <= D; i += 2) {
    if (D % i == 0) {
      primes.push_back(i);
      while (D % i == 0) D /= i;
    }
  }
  if (D > 1) primes.push_back(D);
  cin >> tests;
  for (; tests; --tests) {
    cin >> A >> B;
    long long Q = gcd(A, B);
    long long res = solver(A, Q) * solver(B, Q) % 998244353;
    cout << res << endl;
  }
  return 0;
}

### 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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 500031;
long long D;
vector<long long> primes;
long long gcd(long long a, long long b) {
  while (a && b) a > b ? a %= b : b %= a;
  return a + b;
}
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2) return a * pw(a, b - 1) % 998244353;
  return pw(a * a % 998244353, b / 2);
}
long long inv(long long x) { return pw(x, 998244353 - 2); }
int tests;
long long A, B;
long long ans;
long long fact[N], invf[N];
int count_deg(long long a, long long b) {
  int ret = 0;
  while (a % b == 0) {
    ret++;
    a /= b;
  }
  return ret;
}
long long C(long long n, long long m) {
  long long res = fact[n] * invf[m];
  res %= 998244353;
  res *= invf[n - m];
  return res % 998244353;
}
long long solver(long long a, long long b) {
  long long res = 1;
  long long ttl = 0;
  for (int i = 0; i < primes.size(); i++) {
    int dif = abs(count_deg(a, primes[i]) - count_deg(b, primes[i]));
    res = res * C(ttl + dif, ttl);
    res %= 998244353;
    ttl += dif;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  fact[0] = 1;
  for (int i = 1; i < 100; i++) {
    fact[i] = fact[i - 1] * i % 998244353;
  }
  for (int i = 0; i < 100; i++) {
    invf[i] = inv(fact[i]);
  }
  cin >> D;
  if (D % 2 == 0) {
    primes.push_back(2);
    while (D % 2 == 0) {
      D /= 2;
    }
  }
  for (int i = 3; 1ll * i * i <= D; i += 2) {
    if (D % i == 0) {
      primes.push_back(i);
      while (D % i == 0) D /= i;
    }
  }
  if (D > 1) primes.push_back(D);
  cin >> tests;
  for (; tests; --tests) {
    cin >> A >> B;
    long long Q = gcd(A, B);
    long long res = solver(A, Q) * solver(B, Q) % 998244353;
    cout << res << endl;
  }
  return 0;
}

```