Problem: Iroha has a sequence of N strings s_1, s_2, ..., s_N.

She will choose some (possibly all) strings from the sequence, then concatenate those strings retaining the relative order, to produce a long string.

Among all strings of length K that she can produce in this way, find the lexicographically smallest one.

Constraints

* 1 ≦ N ≦ 2000
* 1 ≦ K ≦ 10^4
* For each i, 1 ≦ |s_i| ≦ K.
* |s_1| + |s_2| + ... + |s_N| ≦ 10^6
* For each i, s_i consists of lowercase letters.
* There exists at least one string of length K that Iroha can produce.

Input

The input is given from Standard Input in the following format:


N K
s_1
s_2
:
s_N


Output

Print the lexicographically smallest string of length K that Iroha can produce.

Examples

Input

3 7
at
coder
codar


Output

atcodar


Input

3 7
coder
codar
at


Output

codarat


Input

4 13
kyuri
namida
zzzzzzz
aaaaaa


Output

namidazzzzzzz

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstring>
#include <bitset>
#include <algorithm>
using namespace std;
typedef pair<int,int> P;
const int maxn=2010;
const int maxk=10010;
bitset<10010> dp[2010];
pair<int,int> vis[2][10010];
char s[2010][10010],ans[1000010];
int len[2010],n,k,cnt,ncnt;
int main() {
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++) scanf("%s",s[i]),len[i]=strlen(s[i]);
    dp[n+1][0]=1;
    for (int i=n;i>=1;i--) 
        dp[i]=dp[i+1]|(dp[i+1]<<len[i]);
    // int cnt=0;
    for (int i=1;i<=n;i++) 
        if (dp[i+1][k-len[i]]) vis[0][++cnt]=make_pair(i,0);

    int ii=0;
    for (int i=0;i<k;i++) {
        char minc='z';
        for (int j=1;j<=cnt;j++) minc=min(minc,s[vis[ii][j].first][vis[ii][j].second]);
        ans[i]=minc;
        
        int minid=n+1;
        ncnt=0;
        for (int j=1;j<=cnt;j++) {
            int nid=vis[ii][j].first,nlen=vis[ii][j].second;
            if (s[nid][nlen]!=minc) continue;
            if (nlen+1==len[nid]) minid=min(minid,nid);
            else vis[ii^1][++ncnt]=make_pair(nid,nlen+1);
        }
        for (int j=minid+1;j<=n;j++) {
            if (k-len[j]-i-1>=0 && dp[j+1][k-len[j]-i-1]) vis[ii^1][++ncnt]=make_pair(j,0);
        }
        ii^=1;
        cnt=ncnt;
    }
    printf("%s\n", ans);
    return 0;
}