#include <bits/stdc++.h>
using namespace std;
long long D, u, v, w, mod = 998244353;
int q;
vector<long long> p;
long long fak[60], faki[60];
long long fpb(long long x, long long y) {
  if (x == 0) return y;
  return fpb(y % x, x);
}
long long pang(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long ret;
  ret = pang(x, y / 2);
  ret *= ret;
  ret %= mod;
  if (y % 2 == 0) {
    return ret;
  } else {
    return (x * ret) % mod;
  }
}
long long f(long long x) {
  if (x == 1) return 1;
  int po, tot = 0;
  vector<long long> v;
  for (long long i : p) {
    if (i > x) {
      if (x > 1) {
        v.push_back(1);
        tot++;
      }
      break;
    }
    po = 0;
    while (x % i == 0) {
      po++;
      x /= i;
    }
    if (po != 0) v.push_back(po);
    tot += po;
  }
  long long ret;
  ret = fak[tot];
  for (long long k : v) {
    ret *= faki[k];
    ret %= mod;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cin >> D >> q;
  int d = sqrt(D);
  for (long long i = 2; i <= d; i++) {
    if (D % i == 0) {
      p.push_back(i);
    }
    while (D % i == 0) {
      D /= i;
    }
  }
  if (D > 1) {
    p.push_back(D);
  }
  long long ans;
  fak[0] = 1;
  for (int i = 1; i <= 50; i++) {
    fak[i] = (fak[i - 1] * i) % mod;
    faki[i] = pang(fak[i], mod - 2);
  }
  while (q--) {
    cin >> u >> v;
    w = fpb(u, v);
    ans = (f(u / w) * f(v / w)) % mod;
    cout << ans << endl;
  }
}

### 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 D, u, v, w, mod = 998244353;
int q;
vector<long long> p;
long long fak[60], faki[60];
long long fpb(long long x, long long y) {
  if (x == 0) return y;
  return fpb(y % x, x);
}
long long pang(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long ret;
  ret = pang(x, y / 2);
  ret *= ret;
  ret %= mod;
  if (y % 2 == 0) {
    return ret;
  } else {
    return (x * ret) % mod;
  }
}
long long f(long long x) {
  if (x == 1) return 1;
  int po, tot = 0;
  vector<long long> v;
  for (long long i : p) {
    if (i > x) {
      if (x > 1) {
        v.push_back(1);
        tot++;
      }
      break;
    }
    po = 0;
    while (x % i == 0) {
      po++;
      x /= i;
    }
    if (po != 0) v.push_back(po);
    tot += po;
  }
  long long ret;
  ret = fak[tot];
  for (long long k : v) {
    ret *= faki[k];
    ret %= mod;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cin >> D >> q;
  int d = sqrt(D);
  for (long long i = 2; i <= d; i++) {
    if (D % i == 0) {
      p.push_back(i);
    }
    while (D % i == 0) {
      D /= i;
    }
  }
  if (D > 1) {
    p.push_back(D);
  }
  long long ans;
  fak[0] = 1;
  for (int i = 1; i <= 50; i++) {
    fak[i] = (fak[i - 1] * i) % mod;
    faki[i] = pang(fak[i], mod - 2);
  }
  while (q--) {
    cin >> u >> v;
    w = fpb(u, v);
    ans = (f(u / w) * f(v / w)) % mod;
    cout << ans << endl;
  }
}

```