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
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2020, maxk = 10100, maxs = 1000100;
using bs = bitset<maxk>;
using ii = pair<int,int>;
bs can[maxn];
string s[maxn];
char ss[maxk];
int len[maxn];
ii q[2][maxs];
int tp[2];
int main(){
	int n, k; scanf("%d %d", &n, &k);
	for(int i = 0; i < n; i++) {
		scanf("%s", ss);
		s[i] = ss;
	}
	for(int i = 0; i < n; i++) len[i] = s[i].size();
	can[n][0] = 1;
	for(int i = n - 1; i >= 0; i--) can[i] = can[i + 1] | (can[i + 1] << len[i]);
	for(int i = 0; i < n; i++) if(k - len[i] >= 0 && can[i + 1][k - len[i]]) q[0][tp[0]++] = {i, 0};
	int cur = 0;
	string res;
	for(int i = 1; i <= k; i++){
		char best = 'z';
		tp[cur] = min(tp[cur], 20000);
		for(int j = 0; j < tp[cur]; j++) best = min(best, s[q[cur][j].first][q[cur][j].second]);
		res.push_back(best);
		int f = n;
		tp[cur ^ 1] = 0;
		for(int j = 0; j < tp[cur]; j++){
			int x = q[cur][j].first, y = q[cur][j].second;
			if(s[x][y] != best) continue;
			if(y + 1 == len[x]) f = min(f, x);
			else q[cur ^ 1][tp[cur ^ 1]++] = {x, y+1};
		}
		for(int j = f + 1; j < n; j++){
			if(k - i - len[j] >= 0 && can[j + 1][k - i - len[j]]){
				q[cur ^ 1][tp[cur ^ 1]++] = {j, 0};
			}
		}
		cur ^= 1;
	}
	printf("%s\n", res.c_str());
	return 0;
}

