#include <bits/stdc++.h>
using namespace std;
const int maxN = 3e5 + 5;
const long long INF = 1e18;
const long long MOD = 998244353;
int fact[maxN], inv[maxN];
long long D;
vector<long long> vec;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
int modP(int a, int b = MOD - 2) {
  if (!b) return 1;
  int c = modP(a, b >> 1);
  c = (1ll * c * c) % MOD;
  if (b & 1) c = (1ll * c * a) % MOD;
  return c;
}
int ans(long long x) {
  int sum = 0;
  int res = 1;
  for (long long d : vec) {
    int cnt = 0;
    while (x % d == 0) {
      x /= d;
      cnt++;
    }
    res = (1ll * res * inv[cnt]) % MOD;
    sum += cnt;
  }
  res = (1ll * res * fact[sum]) % MOD;
  return res;
}
int main() {
  time_t START = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i < maxN; i++) fact[i] = (1ll * i * fact[i - 1]) % MOD;
  inv[maxN - 1] = modP(fact[maxN - 1]);
  for (int i = maxN - 1; i > 0; i--) inv[i - 1] = (1ll * i * inv[i]) % MOD;
  int q;
  scanf("%lld%d", &D, &q);
  for (int i = 2; 1ll * i * i <= D; i++)
    if (!(D % i)) {
      vec.push_back(i);
      while (D % i == 0) D /= i;
    }
  if (D > 1) vec.push_back(D);
  while (q--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    long long g = gcd(u, v);
    u /= g;
    v /= g;
    printf("%lld\n", 1ll * ans(u) * ans(v) % MOD);
  }
  time_t FINISH = clock();
  cerr << "Execution time: "
       << (long double)(FINISH - START) / CLOCKS_PER_SEC * 1000.0
       << " milliseconds.\n";
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 maxN = 3e5 + 5;
const long long INF = 1e18;
const long long MOD = 998244353;
int fact[maxN], inv[maxN];
long long D;
vector<long long> vec;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
int modP(int a, int b = MOD - 2) {
  if (!b) return 1;
  int c = modP(a, b >> 1);
  c = (1ll * c * c) % MOD;
  if (b & 1) c = (1ll * c * a) % MOD;
  return c;
}
int ans(long long x) {
  int sum = 0;
  int res = 1;
  for (long long d : vec) {
    int cnt = 0;
    while (x % d == 0) {
      x /= d;
      cnt++;
    }
    res = (1ll * res * inv[cnt]) % MOD;
    sum += cnt;
  }
  res = (1ll * res * fact[sum]) % MOD;
  return res;
}
int main() {
  time_t START = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i < maxN; i++) fact[i] = (1ll * i * fact[i - 1]) % MOD;
  inv[maxN - 1] = modP(fact[maxN - 1]);
  for (int i = maxN - 1; i > 0; i--) inv[i - 1] = (1ll * i * inv[i]) % MOD;
  int q;
  scanf("%lld%d", &D, &q);
  for (int i = 2; 1ll * i * i <= D; i++)
    if (!(D % i)) {
      vec.push_back(i);
      while (D % i == 0) D /= i;
    }
  if (D > 1) vec.push_back(D);
  while (q--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    long long g = gcd(u, v);
    u /= g;
    v /= g;
    printf("%lld\n", 1ll * ans(u) * ans(v) % MOD);
  }
  time_t FINISH = clock();
  cerr << "Execution time: "
       << (long double)(FINISH - START) / CLOCKS_PER_SEC * 1000.0
       << " milliseconds.\n";
  return 0;
}

```