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>
#define rep(i, a, b) for(int i = (a), i##ed = (b); i <= i##ed; i++)
#define per(i, a, b) for(int i = (a), i##ed = (b); i >= i##ed; i--)
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
int n, k, len[2010], Now = 0;
char s[2010][10010], ans[10010];
bitset<10010> is[2010];
pii vis[2][10010]; int cnt = 0;

template<typename T> void gi(T &x) {
	x = 0; register char c = getchar(), pre = 0;
	for(; c < '0' || c > '9'; pre = c, c = getchar());
	for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10ll + (c ^ 48);
	if(pre == '-') x = -x;
}
int main() {
	gi(n), gi(k);
	rep(i, 1, n) scanf("%s", s[i] + 1), len[i] = strlen(s[i] + 1);
	is[n + 1][0] = 1;
	per(i, n, 1) is[i] = is[i + 1] | (is[i + 1] << len[i]);
	rep(i, 1, n) if(is[i + 1][k - len[i]]) vis[0][++cnt] = mp(i, 1);
	rep(i, 0, k - 1) {
		char c = 'z';
		rep(j, 1, cnt) c = min(c, s[vis[Now][j].fst][vis[Now][j].snd]);
		ans[i] = c; int tmp = 0, id = n + 1;
		rep(j, 1, cnt) {
			int u = vis[Now][j].fst, pos = vis[Now][j].snd;
			if(s[u][pos] != c) continue;
			if(len[u] == pos) id = min(id, u);
			else vis[Now ^ 1][++tmp] = mp(u, pos + 1);
		}
		rep(j, id + 1, n)
			if(k - len[j] - i - 1 >= 0 && is[j + 1][k - len[j] - i - 1])
				vis[Now ^ 1][++tmp] = mp(j, 1);
		Now ^= 1, cnt = tmp;
	}
	rep(i, 0, k - 1) printf("%c", ans[i]);
	return 0;
}