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;
const long long maxn = 1e5 + 5;
bool vis[30];
int main() {
  int n, k;
  memset(vis, 0, sizeof(vis));
  scanf("%d%d", &n, &k);
  string s;
  cin >> s;
  string ans;
  int len = s.size();
  for (int i = 0; i < len; i++) {
    vis[s[i] - 'a'] = 1;
    if (i < k) {
      ans += s[i];
    }
  }
  for (int i = 0; i < 26; i++) {
    if (vis[i]) {
      for (int j = len; j < k; j++) {
        ans += i + 'a';
      }
      break;
    }
  }
  if (k > n) {
    cout << ans << endl;
    return 0;
  }
  int flag = -1;
  for (int i = k - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++) {
      if (vis[j] && j > s[i] - 'a') {
        ans[i] = j + 'a';
        flag = i;
        break;
      }
    }
    if (flag != -1) break;
  }
  for (int i = k - 1; i > flag; i--) {
    for (int j = 0; j < 26; j++) {
      if (vis[j]) {
        ans[i] = j + 'a';
        break;
      }
    }
  }
  cout << ans << endl;
}
