#include <bits/stdc++.h>
using namespace std;
const long long md = 998244353;
long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
long long bin(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long u = bin(a, b / 2);
    return (u * u) % md;
  } else {
    long long u = bin(a, b - 1);
    return (u * a) % md;
  }
}
vector<long long> d;
vector<long long> e;
long long f[1000000 + 5];
long long g[1000000 + 5];
long long solve(long long x, long long y) {
  e.clear();
  long long x0 = x;
  long long y0 = y;
  long long sum = 0;
  for (auto i : d) {
    long long cnt1 = 0;
    long long cnt2 = 0;
    while (x0 % i == 0) {
      cnt1++;
      x0 /= i;
    }
    while (y0 % i == 0) {
      cnt2++;
      y0 /= i;
    }
    if (cnt2 - cnt1 != 0) e.push_back(cnt2 - cnt1);
    sum += cnt2 - cnt1;
  }
  long long ans = f[sum];
  for (auto i : e) {
    ans *= g[i];
    ans %= md;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      d.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) d.push_back(n);
  f[0] = 1;
  for (long long i = 1; i <= 1000; i++) {
    f[i] = f[i - 1] * i;
    f[i] %= md;
  }
  for (long long i = 0; i <= 1000; i++) {
    g[i] = bin(f[i], md - 2);
  }
  long long t;
  cin >> t;
  while (t--) {
    long long x, y;
    cin >> x >> y;
    long long d = gcd(x, y);
    long long ans1 = solve(d, x);
    long long ans2 = solve(d, y);
    long long ans = (ans1 * ans2) % md;
    cout << ans << '\n';
  }
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
const long long md = 998244353;
long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
long long bin(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long u = bin(a, b / 2);
    return (u * u) % md;
  } else {
    long long u = bin(a, b - 1);
    return (u * a) % md;
  }
}
vector<long long> d;
vector<long long> e;
long long f[1000000 + 5];
long long g[1000000 + 5];
long long solve(long long x, long long y) {
  e.clear();
  long long x0 = x;
  long long y0 = y;
  long long sum = 0;
  for (auto i : d) {
    long long cnt1 = 0;
    long long cnt2 = 0;
    while (x0 % i == 0) {
      cnt1++;
      x0 /= i;
    }
    while (y0 % i == 0) {
      cnt2++;
      y0 /= i;
    }
    if (cnt2 - cnt1 != 0) e.push_back(cnt2 - cnt1);
    sum += cnt2 - cnt1;
  }
  long long ans = f[sum];
  for (auto i : e) {
    ans *= g[i];
    ans %= md;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      d.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) d.push_back(n);
  f[0] = 1;
  for (long long i = 1; i <= 1000; i++) {
    f[i] = f[i - 1] * i;
    f[i] %= md;
  }
  for (long long i = 0; i <= 1000; i++) {
    g[i] = bin(f[i], md - 2);
  }
  long long t;
  cin >> t;
  while (t--) {
    long long x, y;
    cin >> x >> y;
    long long d = gcd(x, y);
    long long ans1 = solve(d, x);
    long long ans2 = solve(d, y);
    long long ans = (ans1 * ans2) % md;
    cout << ans << '\n';
  }
}

```