And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples

Input

3 3
abc


Output

aca


Input

3 2
abc


Output

ac


Input

3 3
ayy


Output

yaa


Input

2 3
ba


Output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.
#include <bits/stdc++.h>
using namespace std;
char s[100005], t[100005];
char sets[30], s2[100005];
int cnt = 0, n, k;
void dfs(int len, bool flag) {
  if (len < 0) return;
  if (flag == 1) {
    for (int i = 0; i < cnt; i++) {
      if (s[len] == sets[i]) {
        if (i == cnt - 1) {
          s2[len] = sets[0];
          dfs(len - 1, 1);
        } else {
          s2[len] = sets[i + 1];
          dfs(len - 1, 0);
        }
      }
    }
  } else {
    s2[len] = s[len];
    dfs(len - 1, 0);
  }
}
int main() {
  while (cin >> n >> k) {
    scanf("%s", s);
    for (int i = 0; i < n; i++) s2[i] = s[i];
    sort(s2, s2 + n);
    for (int i = 0; i < n; i++) {
      if (s2[i] == s2[i - 1] && i >= 1) continue;
      sets[cnt++] = s2[i];
    }
    if (k > n) {
      printf("%s", s);
      for (int i = n; i < k; i++) printf("%c", sets[0]);
      puts("");
    } else {
      dfs(k - 1, 1);
      for (int i = 0; i < k; i++) printf("%c", s2[i]);
      puts("");
    }
  }
}
