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 fast_io ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ll long long int
#define f(i,x,n) for(long long int i=x;i<n;i++)
#define fr(i,x,n) for(long long int i=n-1;i>=x;i--)
#define pb push_back
#define mod 1000000007
#define endl '\n'
#define ff first
#define ss second
#define pii pair<int,int>
#define pll pair<ll,ll>
 
 
int main(){
    fast_io
 
    int t; cin >> t;
 
    while(t--){
 
        ll n,a,b; cin >> n >> a >> b;
        
        if(n == 1 || b == 1){
            cout << "YES\n";
            continue;
        }
 
        if(a == 1){
            if((n-1)%b == 0) cout << "YES\n";
            else cout << "NO\n";
 
            continue;
        }
 
        
 
        ll m = 1,  f = 0;
 
        while(m <= n){
            if((n - m)%b == 0){
                f = 1;
                break;
            }
 
            m *= a;
 
        }
 
        if(f) cout << "YES\n";
        else cout << "NO\n";
 
 
 
 
    }
 
    
 
 
    return 0;
}