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>

#define sf(x) scanf("%d", &x)
#define lsf(x) scanf("%lld", &x)
#define pf(x) printf("%d\n", (x))
#define lpf(x) printf("%lld\n",(x))
#define rep(i,x,y) for(int i=x;i<y;i++)  
#define repp(i,x,y) for(int i=x;i<=y;i++)
#define mem(a,x) memset(a,x,sizeof(a))  
#define min(a,b)    (((a) < (b)) ? (a) : (b))
#define max(a,b)    (((a) > (b)) ? (a) : (b))
#define abs(x)    ((x) < 0 ? -(x) : (x))
#define all(x) x.begin(),x.end()
#define sz(x) (x).size()
#define pb push_back
#define x first
#define y second
#define bug(x) cerr<<#x<<'='<<x<<' '
#define debug(x) cerr<<#x<<'='<<x<<endl
#define gcd(a,b) __gcd(a,b)  
#define vec vector<int>

#define mod 1000000007
#define INF 0x3f3f3f3f3f
#define delta 0.85
#define eps 1e-6	
#define PI 3.14159265358979323846
using namespace std;

typedef long long LL;
typedef unsigned long long ULL; 
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const int MAX_N = 2e5 + 10, MAX_M = 2e6 + 10;

void solve() {
		LL n, a, b;
		cin >> n >> a >> b;
		if (a == 1) {
			if ((n - 1)%b == 0) puts("Yes");
			else puts("No");
		}
		else{
			long long res = 1;
			bool flag = 0;
			while(res <= n){
				if((n - res) % b==0) {
					puts("Yes");
					flag = 1;
					return;
				}
				res *= a;
			}
			if(flag == 0)
		puts("NO");
		}
	
}

int main() {
	int t;
 	sf(t); 
	while (t --) {
		solve();
	} 
	return 0;
}