#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
void add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
int binpow(int a, long long n) {
  int res = 1;
  while (n) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
const int N = 100030;
vector<long long> primes;
int fact[N], invfact[N];
vector<int> get(long long v) {
  vector<int> ans(primes.size());
  for (int i = 0; i < ans.size(); i++) {
    while (v % primes[i] == 0) ans[i]++, v /= primes[i];
  }
  return ans;
}
int getans(vector<int> vt, vector<int> nvt) {
  int ans = 1, sum = 0;
  for (int i = 0; i < vt.size(); i++) {
    int deg = nvt[i] - vt[i];
    sum += deg;
    ans = mul(ans, invfact[deg]);
  }
  ans = mul(ans, fact[sum]);
  return ans;
}
void solve() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  invfact[N - 1] = binpow(fact[N - 1], MOD - 2);
  for (int i = N - 2; i >= 0; i--) {
    invfact[i] = mul(invfact[i + 1], i + 1);
  }
  long long D;
  cin >> D;
  for (long long p = 2; p * p <= D; p++) {
    if (D % p == 0) {
      while (D % p == 0) D /= p;
      primes.push_back(p);
    }
  }
  if (D != 1) {
    primes.push_back(D);
  }
  int q;
  cin >> q;
  while (q--) {
    long long v, u;
    cin >> v >> u;
    vector<int> vt1 = get(v), vt2 = get(u);
    vector<int> vt(vt1.size());
    for (int i = 0; i < vt.size(); i++) {
      vt[i] = min(vt1[i], vt2[i]);
    }
    int ans = mul(getans(vt, vt1), getans(vt, vt2));
    cout << ans << "\n";
  }
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  solve();
}

### Prompt

Your task is to create a Cpp solution 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>
using namespace std;
const int MOD = 998244353;
void add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
int binpow(int a, long long n) {
  int res = 1;
  while (n) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
const int N = 100030;
vector<long long> primes;
int fact[N], invfact[N];
vector<int> get(long long v) {
  vector<int> ans(primes.size());
  for (int i = 0; i < ans.size(); i++) {
    while (v % primes[i] == 0) ans[i]++, v /= primes[i];
  }
  return ans;
}
int getans(vector<int> vt, vector<int> nvt) {
  int ans = 1, sum = 0;
  for (int i = 0; i < vt.size(); i++) {
    int deg = nvt[i] - vt[i];
    sum += deg;
    ans = mul(ans, invfact[deg]);
  }
  ans = mul(ans, fact[sum]);
  return ans;
}
void solve() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  invfact[N - 1] = binpow(fact[N - 1], MOD - 2);
  for (int i = N - 2; i >= 0; i--) {
    invfact[i] = mul(invfact[i + 1], i + 1);
  }
  long long D;
  cin >> D;
  for (long long p = 2; p * p <= D; p++) {
    if (D % p == 0) {
      while (D % p == 0) D /= p;
      primes.push_back(p);
    }
  }
  if (D != 1) {
    primes.push_back(D);
  }
  int q;
  cin >> q;
  while (q--) {
    long long v, u;
    cin >> v >> u;
    vector<int> vt1 = get(v), vt2 = get(u);
    vector<int> vt(vt1.size());
    for (int i = 0; i < vt.size(); i++) {
      vt[i] = min(vt1[i], vt2[i]);
    }
    int ans = mul(getans(vt, vt1), getans(vt, vt2));
    cout << ans << "\n";
  }
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  solve();
}

```