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;
int n, k, num;
char s[100050], t[100050];
int p[100050];
map<char, int> m;
int mo[26];
bool vis[26];
bool cmp(char a, char b) { return (a - 'a') < (b - 'a'); }
void sol() {
  cout << s;
  for (int i = n + 1; i <= k; i++) printf("%c", mo[0] + 'a');
  cout << endl;
}
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s);
  num = 0;
  for (int i = 0; i < n; i++) {
    if (!vis[s[i] - 'a']) {
      mo[num++] = s[i] - 'a';
      vis[s[i] - 'a'] = 1;
    }
  }
  sort(mo, mo + num);
  for (int i = 0; i < num; i++) m[mo[i] + 'a'] = i;
  for (int i = 0; i < n; i++) p[i] = m[s[i]];
  p[k - 1]++;
  for (int i = k - 1; i > 0; i--) {
    if (p[i] >= num) {
      p[i] %= num;
      p[i - 1]++;
    }
  }
  if (k > n)
    sol();
  else if (p[0] >= num)
    k++, sol();
  else {
    for (int i = 0; i < k; i++) printf("%c", mo[p[i]] + 'a');
    printf("\n");
  }
  return 0;
}
