#include <bits/stdc++.h>
using namespace std;
long long D, Q, V, U, ans;
vector<long long> primes;
void findPrimeFactors(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      primes.push_back(i);
      while (n != 0 && n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) {
    primes.push_back(n);
  }
}
map<long long, long long> pf[2], gcd;
void findPrimeFactorsByPrimes(long long n, map<long long, long long>& pfs) {
  for (long long i = 0; i < primes.size(); i++) {
    long long cnt = 0;
    while (n != 0 && n % primes[i] == 0) {
      cnt++;
      n /= primes[i];
    }
    if (cnt != 0) {
      pfs[primes[i]] = cnt;
    }
  }
}
vector<long long> inv, rf, f;
void updateInverseModular(int N) {
  inv.resize(N + 1);
  rf.resize(N + 1);
  inv[1] = 1;
  rf[0] = rf[1] = 1;
  for (long long i = 2; i <= N; i++) {
    inv[i] = 998244353 - (998244353 / i) * inv[998244353 % i] % 998244353;
    rf[i] = rf[i - 1] * inv[i] % 998244353;
  }
}
void updateFactorial(int N) {
  f.resize(N + 1);
  f[0] = f[1] = 1;
  for (long long i = 2; i <= N; i++) {
    f[i] = f[i - 1] * i % 998244353;
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  cin >> D;
  findPrimeFactors(D);
  updateFactorial(50);
  updateInverseModular(50);
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    cin >> U >> V;
    ans = 1;
    pf[0].clear();
    pf[1].clear();
    gcd.clear();
    findPrimeFactorsByPrimes(U, pf[0]);
    findPrimeFactorsByPrimes(V, pf[1]);
    for (auto iter = pf[0].begin(); iter != pf[0].end(); iter++) {
      long long num = (*iter).first;
      if (pf[1].count(num) != 0) {
        gcd[num] = pf[0][num] < pf[1][num] ? pf[0][num] : pf[1][num];
      }
    }
    vector<long long> t;
    long long tcnt;
    t.clear();
    tcnt = 0;
    for (auto iter = pf[0].begin(); iter != pf[0].end(); iter++) {
      long long num = (*iter).first, cnt = (*iter).second;
      if (gcd.count(num) != 0) {
        if (cnt > gcd[num]) {
          t.push_back(cnt - gcd[num]);
          tcnt += cnt - gcd[num];
        }
      } else {
        t.push_back(cnt);
        tcnt += cnt;
      }
    }
    ans = (ans * f[tcnt]) % 998244353;
    for (int i = 0; i < t.size(); i++) {
      ans = (ans * rf[t[i]]) % 998244353;
    }
    t.clear();
    tcnt = 0;
    for (auto iter = pf[1].begin(); iter != pf[1].end(); iter++) {
      long long num = (*iter).first, cnt = (*iter).second;
      if (gcd.count(num) != 0) {
        if (cnt > gcd[num]) {
          t.push_back(cnt - gcd[num]);
          tcnt += cnt - gcd[num];
        }
      } else {
        t.push_back(cnt);
        tcnt += cnt;
      }
    }
    ans = (ans * f[tcnt]) % 998244353;
    for (int i = 0; i < t.size(); i++) {
      ans = (ans * rf[t[i]]) % 998244353;
    }
    cout << ans << '\n';
  }
  return 0;
}

### 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 D, Q, V, U, ans;
vector<long long> primes;
void findPrimeFactors(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      primes.push_back(i);
      while (n != 0 && n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) {
    primes.push_back(n);
  }
}
map<long long, long long> pf[2], gcd;
void findPrimeFactorsByPrimes(long long n, map<long long, long long>& pfs) {
  for (long long i = 0; i < primes.size(); i++) {
    long long cnt = 0;
    while (n != 0 && n % primes[i] == 0) {
      cnt++;
      n /= primes[i];
    }
    if (cnt != 0) {
      pfs[primes[i]] = cnt;
    }
  }
}
vector<long long> inv, rf, f;
void updateInverseModular(int N) {
  inv.resize(N + 1);
  rf.resize(N + 1);
  inv[1] = 1;
  rf[0] = rf[1] = 1;
  for (long long i = 2; i <= N; i++) {
    inv[i] = 998244353 - (998244353 / i) * inv[998244353 % i] % 998244353;
    rf[i] = rf[i - 1] * inv[i] % 998244353;
  }
}
void updateFactorial(int N) {
  f.resize(N + 1);
  f[0] = f[1] = 1;
  for (long long i = 2; i <= N; i++) {
    f[i] = f[i - 1] * i % 998244353;
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  cin >> D;
  findPrimeFactors(D);
  updateFactorial(50);
  updateInverseModular(50);
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    cin >> U >> V;
    ans = 1;
    pf[0].clear();
    pf[1].clear();
    gcd.clear();
    findPrimeFactorsByPrimes(U, pf[0]);
    findPrimeFactorsByPrimes(V, pf[1]);
    for (auto iter = pf[0].begin(); iter != pf[0].end(); iter++) {
      long long num = (*iter).first;
      if (pf[1].count(num) != 0) {
        gcd[num] = pf[0][num] < pf[1][num] ? pf[0][num] : pf[1][num];
      }
    }
    vector<long long> t;
    long long tcnt;
    t.clear();
    tcnt = 0;
    for (auto iter = pf[0].begin(); iter != pf[0].end(); iter++) {
      long long num = (*iter).first, cnt = (*iter).second;
      if (gcd.count(num) != 0) {
        if (cnt > gcd[num]) {
          t.push_back(cnt - gcd[num]);
          tcnt += cnt - gcd[num];
        }
      } else {
        t.push_back(cnt);
        tcnt += cnt;
      }
    }
    ans = (ans * f[tcnt]) % 998244353;
    for (int i = 0; i < t.size(); i++) {
      ans = (ans * rf[t[i]]) % 998244353;
    }
    t.clear();
    tcnt = 0;
    for (auto iter = pf[1].begin(); iter != pf[1].end(); iter++) {
      long long num = (*iter).first, cnt = (*iter).second;
      if (gcd.count(num) != 0) {
        if (cnt > gcd[num]) {
          t.push_back(cnt - gcd[num]);
          tcnt += cnt - gcd[num];
        }
      } else {
        t.push_back(cnt);
        tcnt += cnt;
      }
    }
    ans = (ans * f[tcnt]) % 998244353;
    for (int i = 0; i < t.size(); i++) {
      ans = (ans * rf[t[i]]) % 998244353;
    }
    cout << ans << '\n';
  }
  return 0;
}

```