#include <bits/stdc++.h>
using namespace std;
const int SQRTMAX = 4e7;
const int MOD = 998244353;
long long D;
bool d[SQRTMAX + 5];
vector<int> primes;
vector<long long> primesInD;
void PrecomputePrimes() {
  for (int i = 3; i * i <= SQRTMAX; i += 2)
    if (!d[i])
      for (int j = i * i; j <= SQRTMAX; j += 2 * i) d[j] = 1;
  primes.push_back(2);
  for (int i = 3; i <= SQRTMAX; i += 2)
    if (!d[i]) primes.push_back(i);
}
long long GCD(long long x, long long y) {
  long long r = x % y;
  while (r) {
    x = y;
    y = r;
    r = x % y;
  }
  return y;
}
const int DIVMAX = 1e3;
int fact[DIVMAX + 5], invFact[DIVMAX + 5];
int LgPut(int base, int exp) {
  int ans = 1, aux = base;
  for (long long i = 1LL; i <= exp; i <<= 1) {
    if (i & exp) ans = (1LL * ans * aux) % MOD;
    aux = (1LL * aux * aux) % MOD;
  }
  return ans;
}
void PrecFact() {
  fact[0] = 1;
  for (int i = 1; i <= DIVMAX; i++) {
    fact[i] = (1LL * i * fact[i - 1]) % MOD;
    invFact[i] = LgPut(fact[i], MOD - 2);
  }
}
int FactRep(int n, vector<int> vf) {
  int ans = fact[n];
  for (auto it : vf) ans = (1LL * ans * invFact[it]) % MOD;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  PrecomputePrimes();
  PrecFact();
  cin >> D;
  for (auto it : primes)
    if (D % it == 0) {
      primesInD.push_back(it);
      while (D % it == 0) D /= it;
    }
  if (D != 1) primesInD.push_back(D);
  int Q;
  cin >> Q;
  while (Q--) {
    long long x, y;
    cin >> x >> y;
    long long gcd = GCD(x, y);
    long long xx = x / gcd, yy = y / gcd;
    vector<int> frecsX;
    int sumFrecsX = 0;
    vector<int> frecsY;
    int sumFrecsY = 0;
    for (auto it : primesInD)
      if (it <= xx) {
        int exp = 0;
        while (xx % it == 0) {
          exp++;
          xx /= it;
        }
        sumFrecsX += exp;
        if (exp) {
          frecsX.push_back(exp);
        }
      } else
        break;
    if (xx != 1) {
      sumFrecsX += 1;
      frecsX.push_back(1);
    }
    for (auto it : primesInD)
      if (it <= yy) {
        int exp = 0;
        while (yy % it == 0) {
          exp++;
          yy /= it;
        }
        sumFrecsY += exp;
        if (exp) frecsY.push_back(exp);
      } else
        break;
    if (yy != 1) {
      sumFrecsY += 1;
      frecsY.push_back(1);
    }
    cout << (1LL * FactRep(sumFrecsX, frecsX) * FactRep(sumFrecsY, frecsY)) %
                MOD
         << '\n';
  }
  return 0;
}

### 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;
const int SQRTMAX = 4e7;
const int MOD = 998244353;
long long D;
bool d[SQRTMAX + 5];
vector<int> primes;
vector<long long> primesInD;
void PrecomputePrimes() {
  for (int i = 3; i * i <= SQRTMAX; i += 2)
    if (!d[i])
      for (int j = i * i; j <= SQRTMAX; j += 2 * i) d[j] = 1;
  primes.push_back(2);
  for (int i = 3; i <= SQRTMAX; i += 2)
    if (!d[i]) primes.push_back(i);
}
long long GCD(long long x, long long y) {
  long long r = x % y;
  while (r) {
    x = y;
    y = r;
    r = x % y;
  }
  return y;
}
const int DIVMAX = 1e3;
int fact[DIVMAX + 5], invFact[DIVMAX + 5];
int LgPut(int base, int exp) {
  int ans = 1, aux = base;
  for (long long i = 1LL; i <= exp; i <<= 1) {
    if (i & exp) ans = (1LL * ans * aux) % MOD;
    aux = (1LL * aux * aux) % MOD;
  }
  return ans;
}
void PrecFact() {
  fact[0] = 1;
  for (int i = 1; i <= DIVMAX; i++) {
    fact[i] = (1LL * i * fact[i - 1]) % MOD;
    invFact[i] = LgPut(fact[i], MOD - 2);
  }
}
int FactRep(int n, vector<int> vf) {
  int ans = fact[n];
  for (auto it : vf) ans = (1LL * ans * invFact[it]) % MOD;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  PrecomputePrimes();
  PrecFact();
  cin >> D;
  for (auto it : primes)
    if (D % it == 0) {
      primesInD.push_back(it);
      while (D % it == 0) D /= it;
    }
  if (D != 1) primesInD.push_back(D);
  int Q;
  cin >> Q;
  while (Q--) {
    long long x, y;
    cin >> x >> y;
    long long gcd = GCD(x, y);
    long long xx = x / gcd, yy = y / gcd;
    vector<int> frecsX;
    int sumFrecsX = 0;
    vector<int> frecsY;
    int sumFrecsY = 0;
    for (auto it : primesInD)
      if (it <= xx) {
        int exp = 0;
        while (xx % it == 0) {
          exp++;
          xx /= it;
        }
        sumFrecsX += exp;
        if (exp) {
          frecsX.push_back(exp);
        }
      } else
        break;
    if (xx != 1) {
      sumFrecsX += 1;
      frecsX.push_back(1);
    }
    for (auto it : primesInD)
      if (it <= yy) {
        int exp = 0;
        while (yy % it == 0) {
          exp++;
          yy /= it;
        }
        sumFrecsY += exp;
        if (exp) frecsY.push_back(exp);
      } else
        break;
    if (yy != 1) {
      sumFrecsY += 1;
      frecsY.push_back(1);
    }
    cout << (1LL * FactRep(sumFrecsX, frecsX) * FactRep(sumFrecsY, frecsY)) %
                MOD
         << '\n';
  }
  return 0;
}

```