Problem: 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

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

const int kMaxN = 2005, kMaxM = 10005, kMaxS = 1e6 + 5;

int n, k, cnt, len[kMaxN];
char str[kMaxN][kMaxM];
bitset<kMaxM> dp[kMaxN];
pii ch[2][kMaxM];

int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", str[i]);
    len[i] = strlen(str[i]);
  }
  dp[n + 1][0] = 1;
  for (int i = n; i; --i) {
    dp[i] = dp[i + 1] | (dp[i + 1] << len[i]);
  }
  for (int i = 1; i <= n; ++i) {
    if (dp[i + 1][k - len[i]])
      ch[0][cnt++] = make_pair(i, 0);
  }
  for (int now = 1, i = 1; i <= k; ++i, now ^= 1) {
    int ncnt = 0;
    char minc = 'z';
    for (int j = 0; j < cnt; ++j)
      minc = min(minc, str[ch[now ^ 1][j].first][ch[now ^ 1][j].second]);
    putchar(minc);
    int minid = n + 1;
    for (int j = 0; j < cnt; ++j) {
      int id = ch[now ^ 1][j].first, pos = ch[now ^ 1][j].second;
      if (str[id][pos] != minc)
        continue;
      if (pos < len[id] - 1)
        ch[now][ncnt++] = make_pair(id, pos + 1);
      else
        minid = min(minid, id);
    }
    for (int j = minid + 1; j <= n; ++j) {
      if (k - i - len[j] >= 0 && dp[j + 1][k - i - len[j]])
        ch[now][ncnt++] = make_pair(j, 0);
    }
    cnt = ncnt;
  }
  putchar('\n');
  return 0;
}