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;
typedef double db;
typedef long long ll;
typedef pair<int, int> pii;
#define fi first
#define se second
#define lbd(a, x) lower_bound(all(a), x)
const int mod = 1e9 + 7;
const int M = 10100;
const int N = 2100;
bitset<M> ok[N];
int main() {
#ifdef local
	freopen("in.txt", "r", stdin);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int n, k; cin >> n >> k;
	vector<string> vs(n);
	for(auto &s : vs) cin >> s;
	ok[n][0] = 1;
	for(int i = n - 1; i; i--) {
		ok[i] = ok[i + 1] | (ok[i + 1] << vs[i].size());
	}
	vector<pii> q[2];
	int now = 0;
	for(int i = 0; i < n; i++) {
		if(ok[i + 1][k - vs[i].size()]) {
			q[now].emplace_back(i, 0);
		}
	}
	string ans;
	const int S = 10000;
	for(int i = 0; i < k; i++) {
		char z = 'z' + 1;
		for(auto &e : q[now]) {
			z = min(z, vs[e.fi][e.se]);
		}
		ans.push_back(z);
		int zz = n;
		for(auto &e : q[now]) {
			if(vs[e.fi][e.se] == z) {
				if(e.se == vs[e.fi].size() - 1) {
					zz = min(zz, e.fi);
				} else {
					e.se++;
					q[now^1].emplace_back(e);
				}
			}
		}
		int need = k - i - 1;
		for(int i = zz + 1; i < n; i++)
			if(need >= vs[i].size() && ok[i + 1][need - vs[i].size()])
				q[now^1].emplace_back(i, 0);
		q[now].clear();
		if(q[now^1].size() > S) q[now^1].resize(S);
		now ^= 1;
	}
	cout << ans << '\n';
	return 0;
}
