Adding Powers

Suppose you are performing the following algorithm. There is an array v1v1v1,v2v2v2,…,vnvnvn filled with zeroes at start. The following operation is applied to the array several times — at i−thi-thi−th step (000-indexed) you can:

either choose position pos (1≤pos≤n)(1≤pos≤n)(1≤pos≤n) and increase vposv_{pos}vpos by kikiki;
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 vvv equal to the given array a (vj=ajvj=ajvj=aj for each jjj) after some step?

Input

The first line contains one integer T(1≤T≤1000)T (1≤T≤1000)T(1≤T≤1000) — the number of test cases. Next 2T2T2T lines contain test cases — two lines per test case.

The second line contains nnn integers a1,a2,…,an(0≤ai≤1016)a1,a2,…,an (0≤ai≤10^{16})a1,a2,…,an(0≤ai≤1016) — 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.

Example
input
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 k0k0k0 to v1v1v1 and stop the algorithm.

In the third test case, you can’t make two 111 in the array vvv.

In the fifth test case, you can skip 909^090 and 919^191, then add 929^292 and 939^393 to v3v3v3, skip 949^494 and finally, add 959^595 to v2v2v2.

思路

题意：给出一个数列，数字k，数列中的每个数字能否由k的n次方加成，且每个次方只能用一次。很明显将每个数字的k进制表示出来，然后判断同一位次上的数字之和是否 ≤1≤1≤1，这样就可以了，用一个数组统计每一个位次上的和。

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int mx=2e5+10;
int s[100];
ll t,n,k;
ll a[mx];int main()
{scanf("%d",&t);while(t--){bool flag=false;memset(s,0,sizeof(s));scanf("%lld%lld",&n,&k);for(int i=1;i<=n;i++)scanf("%lld",&a[i]);for(int i=1;i<=n;i++){int j=0;while(a[i])//统计每一个数字的k进制{s[j]+=a[i]%k;//加上同一位次的j++;a[i]/=k;}}for(int i=0;i<100;i++)if(s[i]>1)//只要有超过一次的就不能了{flag=true;break;}if(flag)printf("NO\n");elseprintf("YES\n");}return 0;
}

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Adding Powers

Adding Powers

Input

Output

Note

思路

代码

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