#include <bits/stdc++.h>
#pragma ggCC optimize("Ofast")
using namespace std;
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    T t = a % b;
    a = b;
    b = t;
  }
  return a;
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
const long long MOD = 998244353, MAXN = 200042;
long long add(long long a, long long b) { return (a + b) % MOD; }
long long sub(long long a, long long b) { return add(a, MOD - b); }
long long mul(long long a, long long b) { return (a * b) % MOD; }
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2 == 1) z = mul(z, x);
    x = mul(x, x);
    y /= 2;
  }
  return z;
}
long long inv(long long x) { return binpow(x, MOD - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long d, q, u, v;
  cin >> d >> q;
  vector<long long> primes;
  for (long long i = 2; i * i <= d; ++i) {
    if (d % i == 0) {
      primes.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) {
    primes.push_back(d);
  }
  vector<long long> fact(100), rfact(100);
  fact[0] = 1;
  for (long long i = 1; i < 100; ++i) {
    fact[i] = mul(fact[i - 1], i);
    rfact[i] = inv(fact[i]);
  }
  while (q-- > 0) {
    cin >> u >> v;
    vector<long long> up, dw;
    for (auto p : primes) {
      int cnt = 0;
      while (u % p == 0) {
        --cnt;
        u /= p;
      }
      while (v % p == 0) {
        ++cnt;
        v /= p;
      }
      if (cnt < 0)
        dw.push_back(-cnt);
      else if (cnt > 0)
        up.push_back(cnt);
    }
    long long ans = 1;
    ans = mul(ans, fact[accumulate(up.begin(), up.end(), 0)]);
    for (auto it : up) {
      ans = mul(ans, rfact[it]);
    }
    ans = mul(ans, fact[accumulate(dw.begin(), dw.end(), 0)]);
    for (auto it : dw) {
      ans = mul(ans, rfact[it]);
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 ggCC optimize("Ofast")
using namespace std;
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    T t = a % b;
    a = b;
    b = t;
  }
  return a;
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
const long long MOD = 998244353, MAXN = 200042;
long long add(long long a, long long b) { return (a + b) % MOD; }
long long sub(long long a, long long b) { return add(a, MOD - b); }
long long mul(long long a, long long b) { return (a * b) % MOD; }
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2 == 1) z = mul(z, x);
    x = mul(x, x);
    y /= 2;
  }
  return z;
}
long long inv(long long x) { return binpow(x, MOD - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long d, q, u, v;
  cin >> d >> q;
  vector<long long> primes;
  for (long long i = 2; i * i <= d; ++i) {
    if (d % i == 0) {
      primes.push_back(i);
      while (d % i == 0) d /= i;
    }
  }
  if (d > 1) {
    primes.push_back(d);
  }
  vector<long long> fact(100), rfact(100);
  fact[0] = 1;
  for (long long i = 1; i < 100; ++i) {
    fact[i] = mul(fact[i - 1], i);
    rfact[i] = inv(fact[i]);
  }
  while (q-- > 0) {
    cin >> u >> v;
    vector<long long> up, dw;
    for (auto p : primes) {
      int cnt = 0;
      while (u % p == 0) {
        --cnt;
        u /= p;
      }
      while (v % p == 0) {
        ++cnt;
        v /= p;
      }
      if (cnt < 0)
        dw.push_back(-cnt);
      else if (cnt > 0)
        up.push_back(cnt);
    }
    long long ans = 1;
    ans = mul(ans, fact[accumulate(up.begin(), up.end(), 0)]);
    for (auto it : up) {
      ans = mul(ans, rfact[it]);
    }
    ans = mul(ans, fact[accumulate(dw.begin(), dw.end(), 0)]);
    for (auto it : dw) {
      ans = mul(ans, rfact[it]);
    }
    cout << ans << "\n";
  }
  return 0;
}

```