Suppose you are performing the following algorithm. There is an array $v_1, v_2, \dots, v_n$ filled with zeroes at start. The following operation is applied to the array several times — at $i$-th step ($0$-indexed) you can:
- either choose position $pos$ ($1 \le pos \le n$) and increase $v_{pos}$ by $k^i$;
- or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array $v$ equal to the given array $a$ ($v_j = a_j$ for each $j$) after some step?
Input
The first line contains one integer $T$ ($1 \le T \le 1000$) — the number of test cases. Next $2T$ lines contain test cases — two lines per test case.
The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 30$, $2 \le k \le 100$) — the size of arrays $v$ and $a$ and value $k$ used in the algorithm.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{16}$) — the array you’d like to achieve.
Output
For each test case print YES (case insensitive) if you can achieve the array $a$ after some step or NO (case insensitive) otherwise.Exampleinput
5 4 100 0 0 0 0 1 2 1 3 4 1 4 1 3 2 0 1 3 3 9 0 59049 810
output
YES YES NO NO YES
Note
In the first test case, you can stop the algorithm before the $0$-th step, or don’t choose any position several times and stop the algorithm.
In the second test case, you can add $k^0$ to $v_1$ and stop the algorithm.
In the third test case, you can’t make two $1$ in the array $v$.
In the fifth test case, you can skip $9^0$ and $9^1$, then add $9^2$ and $9^3$ to $v_3$, skip $9^4$ and finally, add $9^5$ to $v_2$.
Solution:
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
int _,n,k;
ll a[1010];
bool check() {
set<int> pos;
rep(i,0,n) {
ll x=a[i];
int d=0;
while (x) {
if (x%k!=0&&x%k!=1) {
return 0;
}
if (x%k==1) {
//printf("dd %d\n",d);
if (pos.count(d)) return 0;
pos.insert(d);
}
d+=1;
x/=k;
}
}
return 1;
}
int main() {
for (scanf("%d",&_);_;_--) {
scanf("%d%d",&n,&k);
rep(i,0,n) scanf("%lld",a+i);
puts(check()?"YES":"NO");
}
}
