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
/* --- author: dxm --- */
#include<bits/stdc++.h>
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
typedef long long ll;
inline void Fail(){printf("0");exit(0);}
inline void debug(int x){cerr<<"debug: "<<x<<"\n";}
inline void debug(int n,int x[]){cerr<<"debug "<<"\n";rep(i,n)cerr<<x[i]<<" ";cerr<<"\n";}
inline void debug(string x){cerr<<"debug: "<<x<<"\n";}

const int maxn=2005;
const int maxm=10005;
int n,k;
int C[maxn];
string s[maxn],ans;
bitset<1000005>dp[maxn];
pii cur[1000005][2];
char c[maxn*maxn];

int main(){
	scanf("%d%d",&n,&k);
	rep(i,n){
		scanf("%s",c);
		s[i]=c;
		rep(j,s[i].size())C[s[i][j]-'a']++;
	}
	int t=0;
	char p;
	rep(i,26)if(C[i]>0)t++,p=i+'a';
	if(t==1){
		printf("%s",string(k,p).c_str());
		return 0;
	}
	dp[n][0]=1;
    for(int i=n;i>0;--i)
        dp[i-1]=dp[i]|(dp[i]<<s[i-1].size());
	int cnt=0;
	t=0;
	rep(i,n)if(k>=s[i].size()&&dp[i+1][k-s[i].size()])cur[cnt++][t]=mp(i,0);
	rep(i,k){
		int id=INF;
		char mn='z';
		cnt=min(cnt,10000);
		rep(j,cnt)mn=min(mn,s[cur[j][t].first][cur[j][t].second]);
		ans+=mn;
		int ncnt=0;
		rep(j,cnt){
			int sid=cur[j][t].first,pos=cur[j][t].second;
			if(s[sid][pos]!=mn)continue;
			if(pos+1==s[sid].size())id=min(sid+1,id);
			else cur[ncnt++][t^1]=mp(sid,pos+1);
		}
		for(int j=id;j<n;j++){
			if(k-i-(int)s[j].size()-1>=0)
			if(dp[j+1][k-i-(int)s[j].size()-1])cur[ncnt++][t^1]=mp(j,0);
		}
		cnt=ncnt;t^=1;
	}
	printf("%s",ans.c_str());
	return 0;
}

/*
Input:
dp[i][j]=dp[i+1][j]||dp[i+1][j-s.size()] 
-----------------
Output:
*/