#include <bits/stdc++.h>
using namespace std;
long long int mod = 998244353;
long long int ceil1(long long int n, long long int k) {
  if (n % k) return n / k + 1;
  return n / k;
}
long long int pow1(long long int n, long long int k) {
  long long int ans = 1;
  while (k != 0) {
    if (k % 2) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k = k / 2;
  }
  return ans;
}
bool ispal(string s) {
  long long int i = 0;
  long long int j = s.size() - 1;
  while (i < j) {
    if (s[i] != s[j]) return false;
    i++;
    j--;
  }
  return true;
}
long long int gcdd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcdd(b, a % b);
}
long long int fact(long long int n) {
  if (n == 1 || n == 0)
    return 1;
  else
    return (n * fact(n - 1)) % mod;
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      a.push_back(i);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) a.push_back(n);
  vector<long long int> f(60);
  vector<long long int> inv(60);
  f[0] = 1;
  for (long long int i = 1; i < 60; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
  inv[59] = pow1(f[59], mod - 2);
  for (long long int i = 58; i >= 0; i--) {
    inv[i] = (inv[i + 1] * (i + 1)) % mod;
  }
  long long int q;
  cin >> q;
  while (q--) {
    long long int u, v;
    cin >> u >> v;
    long long int ans1 = 1, ans2 = 1, t1 = 0, t2 = 0;
    for (auto x : a) {
      long long int pa = 0, pb = 0;
      while (u % x == 0) {
        pa++;
        u /= x;
      }
      while (v % x == 0) {
        pb++;
        v /= x;
      }
      if (pa < pb) {
        t1 += pb - pa;
        ans1 = (ans1 * inv[pb - pa]) % mod;
      } else if (pb < pa) {
        t2 += pa - pb;
        ans2 = (ans2 * inv[pa - pb]) % mod;
      }
    }
    ans1 = (ans1 * f[t1]) % mod;
    ans2 = (ans2 * f[t2]) % mod;
    ans1 = (ans1 * ans2) % mod;
    cout << ans1 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int q = 1;
  while (q--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 mod = 998244353;
long long int ceil1(long long int n, long long int k) {
  if (n % k) return n / k + 1;
  return n / k;
}
long long int pow1(long long int n, long long int k) {
  long long int ans = 1;
  while (k != 0) {
    if (k % 2) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k = k / 2;
  }
  return ans;
}
bool ispal(string s) {
  long long int i = 0;
  long long int j = s.size() - 1;
  while (i < j) {
    if (s[i] != s[j]) return false;
    i++;
    j--;
  }
  return true;
}
long long int gcdd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcdd(b, a % b);
}
long long int fact(long long int n) {
  if (n == 1 || n == 0)
    return 1;
  else
    return (n * fact(n - 1)) % mod;
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      a.push_back(i);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) a.push_back(n);
  vector<long long int> f(60);
  vector<long long int> inv(60);
  f[0] = 1;
  for (long long int i = 1; i < 60; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
  inv[59] = pow1(f[59], mod - 2);
  for (long long int i = 58; i >= 0; i--) {
    inv[i] = (inv[i + 1] * (i + 1)) % mod;
  }
  long long int q;
  cin >> q;
  while (q--) {
    long long int u, v;
    cin >> u >> v;
    long long int ans1 = 1, ans2 = 1, t1 = 0, t2 = 0;
    for (auto x : a) {
      long long int pa = 0, pb = 0;
      while (u % x == 0) {
        pa++;
        u /= x;
      }
      while (v % x == 0) {
        pb++;
        v /= x;
      }
      if (pa < pb) {
        t1 += pb - pa;
        ans1 = (ans1 * inv[pb - pa]) % mod;
      } else if (pb < pa) {
        t2 += pa - pb;
        ans2 = (ans2 * inv[pa - pb]) % mod;
      }
    }
    ans1 = (ans1 * f[t1]) % mod;
    ans2 = (ans2 * f[t2]) % mod;
    ans1 = (ans1 * ans2) % mod;
    cout << ans1 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int q = 1;
  while (q--) {
    solve();
  }
  return 0;
}

```