#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
const int INF = 998244353;
long long D;
int Q;
vector<long long> primes;
int fact[70], ifact[70];
int mul(int a, int b) { return (long long)a * b % INF; }
int pwr(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int32_t main() {
  cout << fixed << setprecision(12);
  cerr << fixed << setprecision(4);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  fact[0] = 1;
  for (auto i = (1); i < (69); i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  ifact[68] = pwr(fact[68], INF - 2);
  for (auto i = (68) - 1; i >= (0); i--) {
    ifact[i] = mul(ifact[i + 1], i + 1);
  }
  cin >> D >> Q;
  long long d = D;
  for (long long i = 2; i * i <= D; i++) {
    if (d % i) continue;
    primes.push_back(i);
    while (d % i == 0) d /= i;
  }
  if (d > 1) primes.push_back(d);
  while (Q--) {
    int incr = 0, decr = 0;
    long long u, v;
    cin >> u >> v;
    long long ans = 1;
    for (long long p : primes) {
      int net = 0;
      while (u % p == 0) {
        net++;
        u /= p;
      }
      while (v % p == 0) {
        net--;
        v /= p;
      }
      ans = mul(ans, ifact[abs(net)]);
      if (net < 0)
        incr -= net;
      else
        decr += net;
    }
    ans = mul(ans, fact[incr]);
    ans = mul(ans, fact[decr]);
    cout << ans << '\n';
  }
  return 0;
}

### 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;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
const int INF = 998244353;
long long D;
int Q;
vector<long long> primes;
int fact[70], ifact[70];
int mul(int a, int b) { return (long long)a * b % INF; }
int pwr(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int32_t main() {
  cout << fixed << setprecision(12);
  cerr << fixed << setprecision(4);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  fact[0] = 1;
  for (auto i = (1); i < (69); i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  ifact[68] = pwr(fact[68], INF - 2);
  for (auto i = (68) - 1; i >= (0); i--) {
    ifact[i] = mul(ifact[i + 1], i + 1);
  }
  cin >> D >> Q;
  long long d = D;
  for (long long i = 2; i * i <= D; i++) {
    if (d % i) continue;
    primes.push_back(i);
    while (d % i == 0) d /= i;
  }
  if (d > 1) primes.push_back(d);
  while (Q--) {
    int incr = 0, decr = 0;
    long long u, v;
    cin >> u >> v;
    long long ans = 1;
    for (long long p : primes) {
      int net = 0;
      while (u % p == 0) {
        net++;
        u /= p;
      }
      while (v % p == 0) {
        net--;
        v /= p;
      }
      ans = mul(ans, ifact[abs(net)]);
      if (net < 0)
        incr -= net;
      else
        decr += net;
    }
    ans = mul(ans, fact[incr]);
    ans = mul(ans, fact[decr]);
    cout << ans << '\n';
  }
  return 0;
}

```