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 int maxn = 1e5 + 5;
char s[maxn];
int n, k, vis[30];
vector<char> res;
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) {
    if (vis[s[i] - 'a'] == 0) {
      vis[s[i] - 'a'] = 1;
      res.push_back(s[i]);
    }
  }
  sort(res.begin(), res.end());
  if (k > n) {
    printf("%s", s + 1);
    for (int i = 1; i <= k - n; i++) printf("%c", res[0]);
    puts("");
    return 0;
  } else {
    int pos = -1, flag = 0;
    for (int i = k; i >= 1; i--) {
      if (s[i] != res.back()) {
        pos = i;
        break;
      }
    }
    for (int i = 1; i < pos; i++) printf("%c", s[i]);
    for (int i = 0; i < res.size(); i++)
      if (res[i] > s[pos]) {
        printf("%c", res[i]);
        break;
      }
    for (int i = pos + 1; i <= k; i++) printf("%c", res[0]);
    cout << endl;
  }
}
