#include <bits/stdc++.h>
using namespace std;
constexpr static int MOD = 998244353;
constexpr static int MAXN = 1000;
int64_t D;
vector<pair<int64_t, int>> primes;
map<int64_t, vector<int>> factors;
int64_t inv[MAXN], factorial[MAXN], inv_factorial[MAXN];
void dfs(int64_t val, int i, vector<int>& f) {
  if (i == primes.size()) {
    factors[val] = f;
    return;
  }
  for (int j = 0; j <= primes[i].second; j++) {
    f.push_back(j);
    dfs(val, i + 1, f);
    f.pop_back();
    val *= primes[i].first;
  }
}
void preprocess() {
  for (int64_t i = 2; i * i <= D; i++) {
    if (D % i == 0) {
      D /= i;
      int e = 1;
      while (D % i == 0) D /= i, e++;
      primes.push_back({i, e});
    }
  }
  if (D > 1) primes.push_back({D, 1});
  vector<int> f;
  dfs(1, 0, f);
  inv[1] = 1;
  for (int i = 2; i < MAXN; i++)
    inv[i] = (MOD - (MOD / i) * inv[MOD % i] % MOD) % MOD;
  factorial[0] = 1;
  inv_factorial[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    factorial[i] = factorial[i - 1] * i % MOD;
    inv_factorial[i] = inv_factorial[i - 1] * inv[i] % MOD;
  }
}
int64_t query(int64_t u, int64_t v) {
  int64_t r = 1;
  auto f1 = factors[u];
  auto f2 = factors[v];
  int neg_total = 0;
  int pos_total = 0;
  for (int i = 0; i < f1.size(); i++) {
    int k = f2[i] - f1[i];
    if (k < 0) neg_total += -k;
    if (k > 0) pos_total += k;
    r = r * inv_factorial[abs(k)] % MOD;
  }
  r = r * factorial[neg_total] % MOD;
  r = r * factorial[pos_total] % MOD;
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> D;
  preprocess();
  int Q;
  cin >> Q;
  while (Q--) {
    int64_t u, v;
    cin >> u >> v;
    cout << query(u, v) << '\n';
  }
  cout.flush();
}

### 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;
constexpr static int MOD = 998244353;
constexpr static int MAXN = 1000;
int64_t D;
vector<pair<int64_t, int>> primes;
map<int64_t, vector<int>> factors;
int64_t inv[MAXN], factorial[MAXN], inv_factorial[MAXN];
void dfs(int64_t val, int i, vector<int>& f) {
  if (i == primes.size()) {
    factors[val] = f;
    return;
  }
  for (int j = 0; j <= primes[i].second; j++) {
    f.push_back(j);
    dfs(val, i + 1, f);
    f.pop_back();
    val *= primes[i].first;
  }
}
void preprocess() {
  for (int64_t i = 2; i * i <= D; i++) {
    if (D % i == 0) {
      D /= i;
      int e = 1;
      while (D % i == 0) D /= i, e++;
      primes.push_back({i, e});
    }
  }
  if (D > 1) primes.push_back({D, 1});
  vector<int> f;
  dfs(1, 0, f);
  inv[1] = 1;
  for (int i = 2; i < MAXN; i++)
    inv[i] = (MOD - (MOD / i) * inv[MOD % i] % MOD) % MOD;
  factorial[0] = 1;
  inv_factorial[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    factorial[i] = factorial[i - 1] * i % MOD;
    inv_factorial[i] = inv_factorial[i - 1] * inv[i] % MOD;
  }
}
int64_t query(int64_t u, int64_t v) {
  int64_t r = 1;
  auto f1 = factors[u];
  auto f2 = factors[v];
  int neg_total = 0;
  int pos_total = 0;
  for (int i = 0; i < f1.size(); i++) {
    int k = f2[i] - f1[i];
    if (k < 0) neg_total += -k;
    if (k > 0) pos_total += k;
    r = r * inv_factorial[abs(k)] % MOD;
  }
  r = r * factorial[neg_total] % MOD;
  r = r * factorial[pos_total] % MOD;
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> D;
  preprocess();
  int Q;
  cin >> Q;
  while (Q--) {
    int64_t u, v;
    cin >> u >> v;
    cout << query(u, v) << '\n';
  }
  cout.flush();
}

```