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 inf = 1e9;
string s;
void str_in() {
  s = "";
  char c = getchar();
  while (c < 'a' || c > 'z') {
    c = getchar();
  }
  while (c >= 'a' && c <= 'z') {
    s += c;
    c = getchar();
  }
  return;
}
int n, m;
map<int, char> i_ch;
map<char, int> ch_i;
void swap_string() {
  string s_new = s;
  map<char, bool> cnt;
  for (int i = 0; i < n; i++) {
    cnt[s[i]] = 1;
  }
  map<char, bool>::iterator it;
  int i = 1;
  for (it = cnt.begin(); it != cnt.end(); it++) {
    i_ch[i] = it->first;
    ch_i[it->first] = i;
    i++;
  }
  return;
}
int main() {
  scanf("%d%d", &n, &m);
  str_in();
  swap_string();
  vector<int> ans(n, 0);
  for (int i = 0; i < n; i++) {
    ans[i] = ch_i[s[i]];
  }
  if (m > n) {
    for (int i = 0; i < m; i++) {
      if (i < n) {
        printf("%c", i_ch[ans[i]]);
      } else {
        printf("%c", i_ch[1]);
      }
    }
  } else {
    int pos = 0;
    for (int i = m - 1; i >= 0; i--) {
      if (ans[i] < i_ch.size()) {
        ans[i]++;
        pos = i;
        break;
      }
    }
    for (int i = pos + 1; i < m; i++) {
      ans[i] = 1;
    }
    for (int i = 0; i < m; i++) {
      printf("%c", i_ch[ans[i]]);
    }
  }
  return 0;
}
