#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long inf = 1e17;
const long long md = 998244353;
long long nod(long long a, long long b) { return b == 0 ? a : nod(b, a % b); }
long long n;
map<long long, long long> mp;
vector<pair<long long, long long> > dd;
void build(long long x) {
  for (long long i = 2; i <= x; i++) {
    if (i * i > x) {
      dd.push_back({x, 1});
      break;
    }
    if (x % i == 0) dd.push_back({i, 0});
    while (x / i * i == x) {
      dd.back().second++;
      x /= i;
    }
    if (x == 1) break;
  }
}
void rec1() {
  map<long long, bool> u;
  queue<pair<long long, vector<pair<long long, long long> > > > q;
  q.push({1ll, dd});
  while (!q.empty()) {
    pair<long long, vector<pair<long long, long long> > > cur = q.front();
    q.pop();
    for (long long i = 0; i < cur.second.size(); i++) {
      if (cur.second[i].second == 0) continue;
      if (!u[cur.first * cur.second[i].first]) {
        u[cur.first * cur.second[i].first] = true;
        cur.second[i].second--;
        q.push(make_pair(cur.first * cur.second[i].first, cur.second));
        cur.second[i].second++;
      }
      mp[cur.first * cur.second[i].first] += mp[cur.first];
      if (mp[cur.first * cur.second[i].first] >= md)
        mp[cur.first * cur.second[i].first] -= md;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  build(n);
  mp[1] = 1;
  rec1();
  long long q;
  cin >> q;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long g = nod(u, v);
    u /= g;
    v /= g;
    long long ans = mp[u] * mp[v];
    ans %= md;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Generate 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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long inf = 1e17;
const long long md = 998244353;
long long nod(long long a, long long b) { return b == 0 ? a : nod(b, a % b); }
long long n;
map<long long, long long> mp;
vector<pair<long long, long long> > dd;
void build(long long x) {
  for (long long i = 2; i <= x; i++) {
    if (i * i > x) {
      dd.push_back({x, 1});
      break;
    }
    if (x % i == 0) dd.push_back({i, 0});
    while (x / i * i == x) {
      dd.back().second++;
      x /= i;
    }
    if (x == 1) break;
  }
}
void rec1() {
  map<long long, bool> u;
  queue<pair<long long, vector<pair<long long, long long> > > > q;
  q.push({1ll, dd});
  while (!q.empty()) {
    pair<long long, vector<pair<long long, long long> > > cur = q.front();
    q.pop();
    for (long long i = 0; i < cur.second.size(); i++) {
      if (cur.second[i].second == 0) continue;
      if (!u[cur.first * cur.second[i].first]) {
        u[cur.first * cur.second[i].first] = true;
        cur.second[i].second--;
        q.push(make_pair(cur.first * cur.second[i].first, cur.second));
        cur.second[i].second++;
      }
      mp[cur.first * cur.second[i].first] += mp[cur.first];
      if (mp[cur.first * cur.second[i].first] >= md)
        mp[cur.first * cur.second[i].first] -= md;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  build(n);
  mp[1] = 1;
  rec1();
  long long q;
  cin >> q;
  while (q--) {
    long long u, v;
    cin >> u >> v;
    long long g = nod(u, v);
    u /= g;
    v /= g;
    long long ans = mp[u] * mp[v];
    ans %= md;
    cout << ans << endl;
  }
  return 0;
}

```