Educational Codeforces Round 83 (Rated for Div. 2) C. Adding Powers（数学）

Suppose you are performing the following algorithm. There is an array v1,v2,…,vnv1,v2,…,vn filled with zeroes at start. The following operation is applied to the array several times — at ii -th step (00 -indexed) you can:

• either choose position pospos (1≤pos≤n1≤pos≤n ) and increase vposvpos by kiki ;
• 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 vv equal to the given array aa (vj=ajvj=aj for each jj ) after some step?

Input

The first line contains one integer TT (1≤T≤10001≤T≤1000 ) — the number of test cases. Next 2T2T lines contain test cases — two lines per test case.

The first line of each test case contains two integers nn and kk (1≤n≤301≤n≤30 , 2≤k≤1002≤k≤100 ) — the size of arrays vv and aa and value kk used in the algorithm.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤10160≤ai≤1016 ) — the array you'd like to achieve.

Output

For each test case print YES (case insensitive) if you can achieve the array aa after some step or NO (case insensitive) otherwise.

Example

Input

Copy

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

Copy

YES
YES
NO
NO
YES
大意是给定目标序列，问从起始序列（全0）能否经过若干步操作到达目标序列。对于第i次操作，可以啥都不干也可以给 某个数加上k的i次方。
考虑到可以什么都不干，且每次增加的数彼此互不相同，可以这么想：尝试把目标序列的每个数拆分成若干个k的次方（彼此互不相同）的和的形式（类似二进制拆分）如a[i]=k^p1+k^p2+....k^pn（p1！=p2！=...pn），如果不能成功拆分的话直接就能判定不能到达。然后对于每个数把其p1...pn放进一个vector然后sort，如果有两个数彼此相等则不能到达（每次操作加的数都不一样），反之可以。
详情见注释。

#include <bits/stdc++.h>
using namespace std;
int n,k;
long long ori[35],a[35];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        memset(a,0,sizeof(a));
        int i;
        vector<int>v;
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&ori[i]);
        }
        bool flag=1;
        for(i=1;i<=n;i++)
        {
            long long temp=ori[i];
            int cnt=0;
            if(ori[i]==0)//为0则不需要任何操作 
            {
                continue;
            }
            while(temp)//拆分 
            {
                if(temp%k!=0&&(temp-1)%k!=0)//代表不能拆成k的次方和的形式 
                {
                    flag=0;
                    break;
                }
                if((temp-1)%k==0)//有k的0次方 
                {
                    v.push_back(cnt);//把当前累积的变量扔进去 
                    temp--;//减去1 
                    cnt++;
                    temp/=k;
                }
                else if(temp%k==0)
                {
                    cnt++;
                    temp/=k;
                }        
            }
            if(!flag)break;
        }
        if(!flag)//必须要先判断这个 
        {
            cout<<"NO"<<endl;
            continue;
        }
        if(!v.size())//flag不为0的情况下如果vector是空的 
        {
            cout<<"YES"<<endl;
            continue;
        }
        sort(v.begin(),v.end());
        for(i=0;i<v.size()-1;i++)
        {
            if(v[i]==v[i+1])//判重复 
            {
                flag=0;
                break;
            }
        }
        if(!flag)
        {
            cout<<"NO"<<endl;
            continue;
        }
        else cout<<"YES"<<endl;
    }
    
    
    return 0;
}