#include <bits/stdc++.h>
using namespace std;
const int maxn = 60;
const int mod = 998244353;
long long frac[maxn];
long long inv_frac[maxn];
long long pow(long long a, long long b, int mod) {
  long long ans = 1, base = a;
  while (b) {
    if (b & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    b >>= 1;
  }
  return ans;
}
long long factor[maxn];
int fac_cnt = 0;
long long D;
void init() {
  inv_frac[0] = frac[0] = 1;
  for (int i = 1; i < maxn; i++) {
    frac[i] = (frac[i - 1] * i) % mod;
    inv_frac[i] = pow(frac[i], mod - 2, mod);
  }
  for (long long i = 2; i * i <= D; i++) {
    while (D % i == 0) {
      if (factor[fac_cnt] != i) factor[++fac_cnt] = i;
      D /= i;
    }
  }
  if (D > 1) factor[++fac_cnt] = D;
}
vector<pair<long long, int> > divide(long long n) {
  vector<pair<long long, int> > vec;
  for (long long i = 1; i <= fac_cnt && factor[i] * factor[i] <= n; i++) {
    while (n % factor[i] == 0) {
      if (!vec.empty() && vec.back().first == factor[i])
        vec.back().second++;
      else
        vec.emplace_back(factor[i], 1);
      n /= factor[i];
    }
  }
  if (n != 1) vec.emplace_back(n, 1);
  return vec;
}
long long calc(long long n) {
  auto vec = divide(n);
  int sum = 0;
  for (auto p : vec) sum += p.second;
  long long ans = frac[sum];
  for (auto p : vec) ans = (ans * inv_frac[p.second]) % mod;
  return ans;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  init();
  int Q;
  cin >> D >> Q;
  init();
  for (int i = 1; i <= Q; i++) {
    long long u, v;
    cin >> u >> v;
    if (u < v) swap(u, v);
    long long t = gcd(u, v);
    long long ans = calc(u / t) * calc(v / t) % mod;
    cout << ans << "\n";
  }
}

### 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;
const int maxn = 60;
const int mod = 998244353;
long long frac[maxn];
long long inv_frac[maxn];
long long pow(long long a, long long b, int mod) {
  long long ans = 1, base = a;
  while (b) {
    if (b & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    b >>= 1;
  }
  return ans;
}
long long factor[maxn];
int fac_cnt = 0;
long long D;
void init() {
  inv_frac[0] = frac[0] = 1;
  for (int i = 1; i < maxn; i++) {
    frac[i] = (frac[i - 1] * i) % mod;
    inv_frac[i] = pow(frac[i], mod - 2, mod);
  }
  for (long long i = 2; i * i <= D; i++) {
    while (D % i == 0) {
      if (factor[fac_cnt] != i) factor[++fac_cnt] = i;
      D /= i;
    }
  }
  if (D > 1) factor[++fac_cnt] = D;
}
vector<pair<long long, int> > divide(long long n) {
  vector<pair<long long, int> > vec;
  for (long long i = 1; i <= fac_cnt && factor[i] * factor[i] <= n; i++) {
    while (n % factor[i] == 0) {
      if (!vec.empty() && vec.back().first == factor[i])
        vec.back().second++;
      else
        vec.emplace_back(factor[i], 1);
      n /= factor[i];
    }
  }
  if (n != 1) vec.emplace_back(n, 1);
  return vec;
}
long long calc(long long n) {
  auto vec = divide(n);
  int sum = 0;
  for (auto p : vec) sum += p.second;
  long long ans = frac[sum];
  for (auto p : vec) ans = (ans * inv_frac[p.second]) % mod;
  return ans;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  init();
  int Q;
  cin >> D >> Q;
  init();
  for (int i = 1; i <= Q; i++) {
    long long u, v;
    cin >> u >> v;
    if (u < v) swap(u, v);
    long long t = gcd(u, v);
    long long ans = calc(u / t) * calc(v / t) % mod;
    cout << ans << "\n";
  }
}

```