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;
long long n, k;
char s[100010], t[100010];
int used[30], nxt[30], fir, las;
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s);
  for (int i = 0; i < n; i++) used[s[i] - 'a'] = 1;
  int las = -1;
  for (int i = 0; i < 26; i++)
    if (used[i] == 1) {
      if (las == -1) {
        las = i;
        fir = las;
      } else {
        nxt[las] = i;
        las = i;
      }
    }
  if (n < k) {
    for (int i = 0; i < n; i++) t[i] = s[i];
    for (int i = n; i < k; i++) t[i] = fir + 'a';
  } else {
    for (int i = 0; i < k; i++) t[i] = s[i];
    int j = k - 1;
    while (nxt[t[j] - 'a'] == 0 && j >= 0) {
      t[j] = fir + 'a';
      j--;
    }
    t[j] = nxt[t[j] - 'a'] + 'a';
  }
  printf("%s", t);
}
