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;

string s[2005];
char ans[10005];
int len[2005];
bitset<10005> can[2005];
pair<int,int> rec[2][4005];

#define mp make_pair
#define fi first
#define se second

int main () {
        int n, k; cin >> n >> k;
        for (int i = 1; i <= n; ++i) cin >> s[i], len[i] = s[i].size();
        can[n+1][0] = 1;
        for (int i = n; i >= 1; --i) can[i] = can[i+1] | (can[i+1]<<len[i]);
        int cur = 0, nxt = cur ^ 1, cnt = 0;
        for (int i = 1; i <= n; ++i) if (can[i+1][k-len[i]]) rec[0][++cnt] = mp (i, 0);
        for (int i = 0; i < k; ++i) {
                char Min = 'z';
                for (int j = 1; j <= cnt; ++j) Min = min (Min, s[rec[cur][j].fi][rec[cur][j].se]);
                ans[i] = Min;
                int minID = 1e9, ccnt = 0;
                for (int j = 1; j <= cnt; ++j) {
                        int id = rec[cur][j].fi, pos = rec[cur][j].se;
                        if (s[id][pos] != Min) continue;
                        if (pos+1 == len[id]) minID = min(minID,id);
                        else rec[nxt][++ccnt] = mp (id,pos+1);
                }
                for (int j = minID+1; j <= n; ++j)
                        if (k-len[j]-i-1>=0 && can[j+1][k-len[j]-i-1])
                                rec[nxt][++ccnt] = mp (j,0);    
                cur ^= 1, nxt ^= 1;
                cnt = ccnt;
        }
        ans[k] = '\0';
        cout << ans << endl;
        return 0;
}