Problem: There is an infinite set generated as follows:

  * 1 is in this set. 
  * If x is in this set, x ⋅ a and x+b both are in this set. 



For example, when a=3 and b=6, the five smallest elements of the set are:

  * 1, 
  * 3 (1 is in this set, so 1⋅ a=3 is in this set), 
  * 7 (1 is in this set, so 1+b=7 is in this set), 
  * 9 (3 is in this set, so 3⋅ a=9 is in this set), 
  * 13 (7 is in this set, so 7+b=13 is in this set). 



Given positive integers a, b, n, determine if n is in this set.

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.

The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.

Output

For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.

Example

Input


5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264


Output


Yes
No
Yes
No
Yes

Note

In the first test case, 24 is generated as follows:

  * 1 is in this set, so 3 and 6 are in this set; 
  * 3 is in this set, so 9 and 8 are in this set; 
  * 8 is in this set, so 24 and 13 are in this set. 



Thus we can see 24 is in this set.

The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define int long long 
const int inf = 1e18;
#define D long double
const long double pi = 3.141592653589793238;
#define FIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
const int MOD = 1000000007;
#define fi first
#define se second
#define endl '\n'
const int N = 2e5 + 10;
int n , a , b ; 

void solve(){
    cin >> n >> a >> b ;
    int p = 1;
    if(a==1){
        if((n-a)%b == 0){
            cout << "Yes\n";
        }else{
            cout << "No\n";
        }
        return ;
    }
    while(p<=n){
       if((n-p)%b == 0){
           cout << "Yes\n" ; return ;
       } 
       p *= a;
    }
    cout << "No\n" ;
} 

signed main() {
        FIO
        int t = 1;
        cin >> t;
        while(t--){
            solve();
        }

}