#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int fexp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 998244353;
    b /= 2;
    a = a * a % 998244353;
  }
  return ans;
}
long long int inverse(long long int a, long long int p) {
  return fexp(a, p - 2);
}
void ingraph(vector<vector<long long int> >& graph, long long int m) {
  long long int x, y;
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
}
long long int modify(long long int n) {
  long long int res = n;
  res %= 998244353;
  res += 998244353;
  res %= 998244353;
  return res;
}
vector<long long int> fact(1000001);
vector<long long int> primeFactors(long long int n) {
  vector<long long int> res;
  if (n % 2 == 0) {
    res.push_back(2);
    while (n % 2 == 0) n /= 2;
  }
  for (long long int i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      res.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 2) res.push_back(n);
  return res;
}
void pre() {
  fact[0] = 1;
  for (long long int i = 1; i < 1000001; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= 998244353;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  pre();
  while (t--) {
    long long int n;
    cin >> n;
    long long int q;
    vector<long long int> pf = primeFactors(n);
    sort(pf.begin(), pf.end());
    long long int m = pf.size();
    cin >> q;
    long long int x, y;
    for (long long int i = 0; i < q; i++) {
      cin >> x >> y;
      vector<long long int> a(m), b(m);
      long long int p = x, q = y;
      long long int flag = 0;
      if (x > y) flag = 1;
      for (long long int j = 0; j < m; j++) {
        while (p % pf[j] == 0) {
          a[j]++;
          p /= pf[j];
        }
      }
      for (long long int j = 0; j < m; j++) {
        while (q % pf[j] == 0) {
          b[j]++;
          q /= pf[j];
        }
      }
      vector<long long int> c(m);
      long long int tot = 0;
      long long int tot2 = 0;
      long long int tot1 = 0;
      for (long long int j = 0; j < m; j++) {
        c[j] = abs(a[j] - b[j]);
        if (a[j] > b[j])
          tot1 += c[j];
        else
          tot2 += c[j];
        tot += c[j];
      }
      long long int ans = fact[tot1] * fact[tot2];
      ans %= 998244353;
      for (long long int j = 0; j < m; j++) {
        ans *= inverse(fact[c[j]], 998244353);
        ans %= 998244353;
      }
      cout << ans << "\n";
    }
  }
}

### 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>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int fexp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 998244353;
    b /= 2;
    a = a * a % 998244353;
  }
  return ans;
}
long long int inverse(long long int a, long long int p) {
  return fexp(a, p - 2);
}
void ingraph(vector<vector<long long int> >& graph, long long int m) {
  long long int x, y;
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
}
long long int modify(long long int n) {
  long long int res = n;
  res %= 998244353;
  res += 998244353;
  res %= 998244353;
  return res;
}
vector<long long int> fact(1000001);
vector<long long int> primeFactors(long long int n) {
  vector<long long int> res;
  if (n % 2 == 0) {
    res.push_back(2);
    while (n % 2 == 0) n /= 2;
  }
  for (long long int i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      res.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 2) res.push_back(n);
  return res;
}
void pre() {
  fact[0] = 1;
  for (long long int i = 1; i < 1000001; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= 998244353;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  pre();
  while (t--) {
    long long int n;
    cin >> n;
    long long int q;
    vector<long long int> pf = primeFactors(n);
    sort(pf.begin(), pf.end());
    long long int m = pf.size();
    cin >> q;
    long long int x, y;
    for (long long int i = 0; i < q; i++) {
      cin >> x >> y;
      vector<long long int> a(m), b(m);
      long long int p = x, q = y;
      long long int flag = 0;
      if (x > y) flag = 1;
      for (long long int j = 0; j < m; j++) {
        while (p % pf[j] == 0) {
          a[j]++;
          p /= pf[j];
        }
      }
      for (long long int j = 0; j < m; j++) {
        while (q % pf[j] == 0) {
          b[j]++;
          q /= pf[j];
        }
      }
      vector<long long int> c(m);
      long long int tot = 0;
      long long int tot2 = 0;
      long long int tot1 = 0;
      for (long long int j = 0; j < m; j++) {
        c[j] = abs(a[j] - b[j]);
        if (a[j] > b[j])
          tot1 += c[j];
        else
          tot2 += c[j];
        tot += c[j];
      }
      long long int ans = fact[tot1] * fact[tot2];
      ans %= 998244353;
      for (long long int j = 0; j < m; j++) {
        ans *= inverse(fact[c[j]], 998244353);
        ans %= 998244353;
      }
      cout << ans << "\n";
    }
  }
}

```