#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
using namespace std;
const long long MOD = 998244353;
mt19937_64 rnd(time(0));
long long n, m, k, d, h, g, T, Q, kol[1000001];
vector<long long> v;
map<long long, int> w;
bool tt[1000001];
long long binpow(long long h, long long r, long long md = MOD) {
  long long l = 1;
  while (r) {
    if (r & 1) l *= h, l %= md;
    h *= h;
    h %= md;
    r /= 2;
  }
  return l;
}
long long gcd(long long x, long long y) {
  if (x < y) swap(x, y);
  while (x && y) {
    x %= y;
    swap(x, y);
  }
  return x + y;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long D;
  cin >> D;
  long long p = sqrt(D);
  for (long long i = 1; i <= p; i++)
    if (D % i == 0) {
      v.push_back(i);
      if (i != D / i) v.push_back(D / i);
    }
  sort(v.begin(), v.end());
  long long n = v.size() - 1;
  vector<long long> smp;
  w[1] = 0;
  for (int i = 1; i <= n; i++) {
    w[v[i]] = i;
    if (!tt[i]) {
      smp.push_back(v[i]);
      for (int j = i + 1; j <= n; j++)
        if (v[j] % v[i] == 0) {
          tt[j] = 1;
        }
    }
  }
  kol[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < smp.size(); j++)
      if (v[i] % smp[j] == 0) {
        kol[i] = (kol[i] + kol[w[v[i] / smp[j]]]) % MOD;
      }
  }
  long long q;
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    long long z = gcd(x, y);
    x /= z;
    y /= z;
    cout << (kol[w[x]] * kol[w[y]]) % 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>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
using namespace std;
const long long MOD = 998244353;
mt19937_64 rnd(time(0));
long long n, m, k, d, h, g, T, Q, kol[1000001];
vector<long long> v;
map<long long, int> w;
bool tt[1000001];
long long binpow(long long h, long long r, long long md = MOD) {
  long long l = 1;
  while (r) {
    if (r & 1) l *= h, l %= md;
    h *= h;
    h %= md;
    r /= 2;
  }
  return l;
}
long long gcd(long long x, long long y) {
  if (x < y) swap(x, y);
  while (x && y) {
    x %= y;
    swap(x, y);
  }
  return x + y;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long D;
  cin >> D;
  long long p = sqrt(D);
  for (long long i = 1; i <= p; i++)
    if (D % i == 0) {
      v.push_back(i);
      if (i != D / i) v.push_back(D / i);
    }
  sort(v.begin(), v.end());
  long long n = v.size() - 1;
  vector<long long> smp;
  w[1] = 0;
  for (int i = 1; i <= n; i++) {
    w[v[i]] = i;
    if (!tt[i]) {
      smp.push_back(v[i]);
      for (int j = i + 1; j <= n; j++)
        if (v[j] % v[i] == 0) {
          tt[j] = 1;
        }
    }
  }
  kol[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < smp.size(); j++)
      if (v[i] % smp[j] == 0) {
        kol[i] = (kol[i] + kol[w[v[i] / smp[j]]]) % MOD;
      }
  }
  long long q;
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    long long z = gcd(x, y);
    x /= z;
    y /= z;
    cout << (kol[w[x]] * kol[w[y]]) % MOD << "\n";
  }
  return 0;
}

```