#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
const int mod = 998244353;
const long long INF = 1e18;
map<long long, long long> mp;
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b)
    d = a, x = 1, y = 0;
  else
    exgcd(b, a % b, d, y, x), y -= x * (a / b);
}
long long C(long long n, long long m, long long p) {
  if (m > n) return 0;
  if (m > n - m) m = n - m;
  long long ans = 1, tmp = 1, g, x, y;
  for (int i = 0; i < m; i++) ans = ans * (n - i) % p, tmp = tmp * (i + 1) % p;
  exgcd(tmp, p, g, x, y);
  return (ans * (x % p) % p + p) % p;
}
long long cal(long long x) {
  vector<long long> V;
  for (long long i = 2; i * i <= x; i++) {
    long long res = 0;
    while (x % i == 0) x /= i, res++;
    if (res) V.push_back(res);
  }
  if (x > 1) V.push_back(1);
  long long res = 0, ans = 1;
  for (auto i : V) {
    res += i;
    ans = ans * C(res, i, mod) % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void rua() {
  long long d, q;
  scanf("%lld%lld", &d, &q);
  while (q--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    long long tmp = gcd(u, v);
    if (!mp.count(u / tmp)) mp[u / tmp] = cal(u / tmp);
    if (!mp.count(v / tmp)) mp[v / tmp] = cal(v / tmp);
    printf("%lld\n", mp[u / tmp] * mp[v / tmp] % mod);
  }
}
int main() {
  rua();
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxn = 2e5 + 7;
const int mod = 998244353;
const long long INF = 1e18;
map<long long, long long> mp;
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b)
    d = a, x = 1, y = 0;
  else
    exgcd(b, a % b, d, y, x), y -= x * (a / b);
}
long long C(long long n, long long m, long long p) {
  if (m > n) return 0;
  if (m > n - m) m = n - m;
  long long ans = 1, tmp = 1, g, x, y;
  for (int i = 0; i < m; i++) ans = ans * (n - i) % p, tmp = tmp * (i + 1) % p;
  exgcd(tmp, p, g, x, y);
  return (ans * (x % p) % p + p) % p;
}
long long cal(long long x) {
  vector<long long> V;
  for (long long i = 2; i * i <= x; i++) {
    long long res = 0;
    while (x % i == 0) x /= i, res++;
    if (res) V.push_back(res);
  }
  if (x > 1) V.push_back(1);
  long long res = 0, ans = 1;
  for (auto i : V) {
    res += i;
    ans = ans * C(res, i, mod) % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void rua() {
  long long d, q;
  scanf("%lld%lld", &d, &q);
  while (q--) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    long long tmp = gcd(u, v);
    if (!mp.count(u / tmp)) mp[u / tmp] = cal(u / tmp);
    if (!mp.count(v / tmp)) mp[v / tmp] = cal(v / tmp);
    printf("%lld\n", mp[u / tmp] * mp[v / tmp] % mod);
  }
}
int main() {
  rua();
  return 0;
}

```