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
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
//#pragma GCC target("avx,tune=native")
#include<bits/stdc++.h>
//#include<ext/pb_ds/assoc_container.hpp>
using namespace std;
//using namespace __gnu_pbds;
const int inf=0x3f3f3f3f;
const double eps=1e-6;
const int mod=1000000007;
typedef long long ll;
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
int modpow(int x,int y,int md=mod){if(y==0)return 1;int ret=modpow(x,y>>1,md);ret=(ll)ret*ret%md;if(y&1)ret=(ll)ret*x%md;return ret;}
bitset<1000005>state,segtree[66];
bitset<10005>dp[2020];
int pos[2020];
string s[2020];
int n,k;
char get(){
	int now=1;
	while(now<32){
		if((state&segtree[now<<1]).none())now=now*2+1;
		else now*=2;
	}
	return 'a'+now-32;
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>k;
	for(int i=0;i<n;i++)cin>>s[i];
	dp[n].set(0);
	for(int i=n-1;i>=0;i--){
		dp[i]=dp[i+1]|(dp[i+1]<<((int)s[i].size()));
	}
	string tmp="";
	for(int i=0;i<n;i++){
		pos[i]=tmp.size();
		tmp+=s[i];
	}
	pos[n]=tmp.size();
	for(int i=0;i<64;i++)segtree[i].reset();
	for(int i=0;i<tmp.size();i++){
		segtree[tmp[i]-'a'+32].set(i);
	}
	for(int i=31;i>=1;i--)segtree[i]=segtree[i<<1]|segtree[i<<1|1];
	state.reset();
	state[0]=1;
	string ans="";
	for(int i=0;i<k;i++){
		for(int j=0;j<n;j++)if(state[pos[j]])state.set(pos[j+1]);
		for(int j=0;j<n;j++)if(k-i-(int)s[j].size()<0||!dp[j+1][k-i-(int)s[j].size()])state.reset(pos[j]);
		char c=get();
		ans.push_back(c);
		state&=segtree[32+c-'a'];
		state<<=1;
	}
	cout<<ans<<endl;
	return 0;
}
