#include <bits/stdc++.h>
using namespace std;
long long mod = (998244353LL);
inline long long Mod(long long a, long long b) { return (a % b); }
inline long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b, swap(a, b);
  }
  return a;
}
void read(vector<long long> &w, long long n) {
  w.resize(n);
  for (long long i = 0; i < n; i++) cin >> w[i];
}
void print(vector<long long> &w) {
  for (long long i = 0; i < (long long)(w).size(); i++) {
    if (i == (long long)(w).size() - 1)
      cout << w[i] << "\n";
    else
      cout << w[i] << " ";
  }
}
long long n, m, q, k, v[300050], ans;
pair<long long, long long> w[300050];
string second;
int32_t main() {
  cin >> n >> q;
  vector<long long> primes, fat(500), inv(500);
  fat[0] = 1LL;
  inv[0] = 1LL;
  for (long long i = 1; i < 500; i++) {
    fat[i] = (i * fat[i - 1]) % mod;
    inv[i] = poww(fat[i], mod - 2);
  }
  for (long long i = 2; i * i <= n; i++) {
    long long cnt = 0;
    while (n % i == 0) {
      cnt++;
      n /= i;
    }
    if (cnt > 0) primes.push_back(i);
  }
  if (n > 1) primes.push_back(n);
  while (q--) {
    long long a, b;
    cin >> a >> b;
    long long x = gcd(a, b);
    long long A = a / x, B = b / x, ans = 1LL;
    long long tota = 0, totb = 0;
    for (auto p : primes) {
      long long qa = 0, qb = 0;
      while (A % p == 0) {
        qa++;
        A /= p;
      }
      while (B % p == 0) {
        qb++;
        B /= p;
      }
      tota += qa;
      totb += qb;
      long long inA = inv[qa];
      long long inB = inv[qb];
      ans = Mod(inB * Mod(inA * ans, mod), mod);
    }
    ans = Mod(fat[totb] * Mod(ans * fat[tota], mod), mod);
    cout << ans << "\n";
  }
}

### Prompt

Your task is to create a cpp solution 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 mod = (998244353LL);
inline long long Mod(long long a, long long b) { return (a % b); }
inline long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b, swap(a, b);
  }
  return a;
}
void read(vector<long long> &w, long long n) {
  w.resize(n);
  for (long long i = 0; i < n; i++) cin >> w[i];
}
void print(vector<long long> &w) {
  for (long long i = 0; i < (long long)(w).size(); i++) {
    if (i == (long long)(w).size() - 1)
      cout << w[i] << "\n";
    else
      cout << w[i] << " ";
  }
}
long long n, m, q, k, v[300050], ans;
pair<long long, long long> w[300050];
string second;
int32_t main() {
  cin >> n >> q;
  vector<long long> primes, fat(500), inv(500);
  fat[0] = 1LL;
  inv[0] = 1LL;
  for (long long i = 1; i < 500; i++) {
    fat[i] = (i * fat[i - 1]) % mod;
    inv[i] = poww(fat[i], mod - 2);
  }
  for (long long i = 2; i * i <= n; i++) {
    long long cnt = 0;
    while (n % i == 0) {
      cnt++;
      n /= i;
    }
    if (cnt > 0) primes.push_back(i);
  }
  if (n > 1) primes.push_back(n);
  while (q--) {
    long long a, b;
    cin >> a >> b;
    long long x = gcd(a, b);
    long long A = a / x, B = b / x, ans = 1LL;
    long long tota = 0, totb = 0;
    for (auto p : primes) {
      long long qa = 0, qb = 0;
      while (A % p == 0) {
        qa++;
        A /= p;
      }
      while (B % p == 0) {
        qb++;
        B /= p;
      }
      tota += qa;
      totb += qb;
      long long inA = inv[qa];
      long long inB = inv[qb];
      ans = Mod(inB * Mod(inA * ans, mod), mod);
    }
    ans = Mod(fat[totb] * Mod(ans * fat[tota], mod), mod);
    cout << ans << "\n";
  }
}

```