#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 998244353;
ll pw(ll a, ll n) {
  ll ret = 1;
  while (n > 0) {
    if (n & 1 == 1) {
      ret = (ret * a) % MOD;
    }
    a = (a * a) % MOD;
    n = n >> 1;
  }
  return ret;
}
ll inv(ll a) { return pw(a, MOD - 2); }
void solve() {
  ll D;
  cin >> D;
  vl primes;
  for (ll i = 2; i * i <= D; i++) {
    if (D % i == 0) {
      primes.push_back(i);
      while (D % i == 0) {
        D /= i;
      }
    }
  }
  if (D > 1) {
    primes.push_back(D);
  }
  vl fat(101);
  fat[0] = 1;
  fat[1] = 1;
  vl invfat(101);
  invfat[0] = 1;
  invfat[1] = 1;
  for (int i = 2; i <= 100; i++) {
    fat[i] = (fat[i - 1] * i) % MOD;
    invfat[i] = (invfat[i - 1] * inv(i)) % MOD;
  }
  int sz = primes.size();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    ll u, v;
    cin >> u >> v;
    vl difs_u;
    vl difs_d;
    ll sum1 = 0;
    ll sum2 = 0;
    ll ans = 1;
    for (int i = 0; i < sz; i++) {
      ll x1 = 0;
      ll x2 = 0;
      while (u % primes[i] == 0) {
        x1++;
        u /= primes[i];
      }
      while (v % primes[i] == 0) {
        x2++;
        v /= primes[i];
      }
      ll dif = x1 - x2;
      if (dif == 0) continue;
      if (dif < 0) {
        dif = -dif;
        sum2 += dif;
      } else {
        sum1 += dif;
      }
      ans = (ans * invfat[dif]) % MOD;
    }
    ans = (ans * fat[sum1]) % MOD;
    ans = (ans * fat[sum2]) % MOD;
    cout << ans << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 998244353;
ll pw(ll a, ll n) {
  ll ret = 1;
  while (n > 0) {
    if (n & 1 == 1) {
      ret = (ret * a) % MOD;
    }
    a = (a * a) % MOD;
    n = n >> 1;
  }
  return ret;
}
ll inv(ll a) { return pw(a, MOD - 2); }
void solve() {
  ll D;
  cin >> D;
  vl primes;
  for (ll i = 2; i * i <= D; i++) {
    if (D % i == 0) {
      primes.push_back(i);
      while (D % i == 0) {
        D /= i;
      }
    }
  }
  if (D > 1) {
    primes.push_back(D);
  }
  vl fat(101);
  fat[0] = 1;
  fat[1] = 1;
  vl invfat(101);
  invfat[0] = 1;
  invfat[1] = 1;
  for (int i = 2; i <= 100; i++) {
    fat[i] = (fat[i - 1] * i) % MOD;
    invfat[i] = (invfat[i - 1] * inv(i)) % MOD;
  }
  int sz = primes.size();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    ll u, v;
    cin >> u >> v;
    vl difs_u;
    vl difs_d;
    ll sum1 = 0;
    ll sum2 = 0;
    ll ans = 1;
    for (int i = 0; i < sz; i++) {
      ll x1 = 0;
      ll x2 = 0;
      while (u % primes[i] == 0) {
        x1++;
        u /= primes[i];
      }
      while (v % primes[i] == 0) {
        x2++;
        v /= primes[i];
      }
      ll dif = x1 - x2;
      if (dif == 0) continue;
      if (dif < 0) {
        dif = -dif;
        sum2 += dif;
      } else {
        sum1 += dif;
      }
      ans = (ans * invfat[dif]) % MOD;
    }
    ans = (ans * fat[sum1]) % MOD;
    ans = (ans * fat[sum2]) % MOD;
    cout << ans << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

```