Codeforces Round #638 C. Phoenix and Distribution（字典序/排序/构造）

Phoenix has a string ss consisting of lowercase Latin letters. He wants to distribute all the letters of his string into kk non-empty strings a1,a2,…,aka1,a2,…,ak such that every letter of ss goes to exactly one of the strings aiai . The strings aiai do not need to be substrings of ss . Phoenix can distribute letters of ss and rearrange the letters within each string aiai however he wants.

For example, if s=s= baba and k=2k=2 , Phoenix may distribute the letters of his string in many ways, such as:

• ba and ba
• a and abb
• ab and ab
• aa and bb

But these ways are invalid:

• baa and ba
• b and ba
• baba and empty string (aiai should be non-empty)

Phoenix wants to distribute the letters of his string ss into kk strings a1,a2,…,aka1,a2,…,ak to minimize the lexicographically maximum string among them, i. e. minimize max(a1,a2,…,ak)max(a1,a2,…,ak) . Help him find the optimal distribution and print the minimal possible value of max(a1,a2,…,ak)max(a1,a2,…,ak) .

String xx is lexicographically less than string yy if either xx is a prefix of yy and x≠yx≠y , or there exists an index ii (1≤i≤min(|x|,|y|))1≤i≤min(|x|,|y|)) such that xixi < yiyi and for every jj (1≤j<i)(1≤j<i) xj=yjxj=yj . Here |x||x| denotes the length of the string xx .

Input

The input consists of multiple test cases. The first line contains an integer tt (1≤t≤10001≤t≤1000 ) — the number of test cases. Each test case consists of two lines.

The first line of each test case consists of two integers nn and kk (1≤k≤n≤1051≤k≤n≤105 ) — the length of string ss and the number of non-empty strings, into which Phoenix wants to distribute letters of ss , respectively.

The second line of each test case contains a string ss of length nn consisting only of lowercase Latin letters.

It is guaranteed that the sum of nn over all test cases is ≤105≤105 .

Output

Print tt answers — one per test case. The ii -th answer should be the minimal possible value of max(a1,a2,…,ak)max(a1,a2,…,ak) in the ii -th test case.

Example

Input

Copy

6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix

Output

Copy

ab
abbc
b
aa
x
ehinopx

大意就是给一个字符串，可以把字母顺序打乱，然后用这些字母构造k个串使得字典序最大的字符串的字典序最小。
首先可以想到，因为这k个串非空，所以都必然有一个首字母，把原字符串从小到大sort后的第k个字母一定是最终造出来的字典序最大的字符串的首字母。然后这么看，假设第k个字母是x，如果第1~k-1里没有x或者不全是x的话，直接输出一个x即可，因为可以把剩下的所有字母都填到最小的字母后面，构成类似ayyyzzz b c d ... x这样，x的字典序大于ayyyzzz；如果1~k个字母全是x的话，考虑第k+1到第n个字母，如果这些字母全部相同的话，则平均分到每个字符串，如果不全部相同，则把剩下的字母全部添加到一个x的后面（假设不全部添加到某个x后面的话，会发现不是最优情况，比如wxyz和w以及wyz和wx，肯定是前者更优）。

#include <bits/stdc++.h>
using namespace std;
char s[100005];
int n,k,i;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        scanf("%s",s);
        vector<char>v;
        sort(s,s+n);
        v.push_back(s[k-1]);
        int cnt=1;
        if(k==1)
        {
            cout<<s<<endl;
            continue;
        }
        for(i=0;i<n;i++)
        {
            if(i<k-1&&s[i]==s[k-1])cnt++;
        }
        bool flag=1;
        for(i=k;i<n;i++)
        {
            if(s[i]!=s[k])flag=0;
        }
        if(cnt!=k)//剩下的全给最小的那个 
        {
            cout<<s[k-1]<<endl;
            continue;
        }
        else
        {
            if(!flag)
            {
                for(i=k;i<n;i++)v.push_back(s[i]); 
            }
            else
            {
                int res=(n-k)%k;
                for(i=1;i<=(n-k)/k;i++)v.push_back(s[k]);
                if(res)v.push_back(s[k]);
            }
        }
        for(i=0;i<v.size();i++)
        {
            cout<<v[i];
        }
        cout<<endl;
    }
}