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 + 100;
int k, n;
char s[maxn], t[maxn];
int dp[maxn][26], vis[26], flag = 0;
;
void dfs(int pos, int pre, bool limt) {
  if (flag) return;
  if (pos == k) {
    if (!limt) {
      flag = 1;
      for (int i = 0; i < k; ++i) printf("%c", t[i]);
      puts("");
    }
    return;
  }
  int top = limt ? s[pos] - 'a' : 0;
  if (limt && pos >= n) top = 0;
  for (int i = top; i < 26; ++i) {
    if (!vis[i]) continue;
    t[pos] = i + 'a';
    dfs(pos + 1, i, limt && i == s[pos] - 'a' && pos < n);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s);
  for (int i = 0; i < n; ++i) {
    vis[s[i] - 'a']++;
  }
  dfs(0, 0, 1);
  return 0;
}
