#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e5 + 10;
const ll M = 998244353;
ll d;
int q;
vector<ll> dpd, fact, afact;
ll gcd(ll a, ll b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
ll pw(ll a, ll n) {
  if (n == 0) return 1;
  if (n & 1) return a * pw(a, n - 1) % M;
  ll temp = pw(a, n >> 1);
  return temp * temp % M;
}
void count_dpd() {
  ll x = d;
  for (ll i = 2; i * i <= x; i++) {
    if (x % i == 0) dpd.push_back(i);
    while (x % i == 0) x /= i;
  }
  if (x > 1) dpd.push_back(x);
}
void count_fact() {
  fact.push_back(1);
  afact.push_back(1);
  for (int i = 1; i < N; i++) {
    fact.push_back((fact.back() * i) % M);
    afact.push_back((afact.back() * pw(i, M - 2)) % M);
  }
}
ll answer(ll a) {
  vector<int> ks;
  int k = 0;
  for (ll x : dpd) {
    ks.push_back(0);
    while (a % x == 0) {
      ks.back()++;
      a /= x;
    }
    k += ks.back();
    if (ks.back() == 0) ks.pop_back();
  }
  ll ans = fact[k];
  for (ll x : ks) ans = (ans * afact[x]) % M;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> d >> q;
  count_dpd();
  count_fact();
  while (q--) {
    ll v, u;
    cin >> v >> u;
    ll g = gcd(v, u);
    cout << (answer(v / g) * answer(u / g)) % M << '\n';
  }
  return 0;
}

### 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;
using ll = long long;
const int N = 1e5 + 10;
const ll M = 998244353;
ll d;
int q;
vector<ll> dpd, fact, afact;
ll gcd(ll a, ll b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
ll pw(ll a, ll n) {
  if (n == 0) return 1;
  if (n & 1) return a * pw(a, n - 1) % M;
  ll temp = pw(a, n >> 1);
  return temp * temp % M;
}
void count_dpd() {
  ll x = d;
  for (ll i = 2; i * i <= x; i++) {
    if (x % i == 0) dpd.push_back(i);
    while (x % i == 0) x /= i;
  }
  if (x > 1) dpd.push_back(x);
}
void count_fact() {
  fact.push_back(1);
  afact.push_back(1);
  for (int i = 1; i < N; i++) {
    fact.push_back((fact.back() * i) % M);
    afact.push_back((afact.back() * pw(i, M - 2)) % M);
  }
}
ll answer(ll a) {
  vector<int> ks;
  int k = 0;
  for (ll x : dpd) {
    ks.push_back(0);
    while (a % x == 0) {
      ks.back()++;
      a /= x;
    }
    k += ks.back();
    if (ks.back() == 0) ks.pop_back();
  }
  ll ans = fact[k];
  for (ll x : ks) ans = (ans * afact[x]) % M;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> d >> q;
  count_dpd();
  count_fact();
  while (q--) {
    ll v, u;
    cin >> v >> u;
    ll g = gcd(v, u);
    cout << (answer(v / g) * answer(u / g)) % M << '\n';
  }
  return 0;
}

```