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>
const int MaxN = 1e5;
using namespace std;
int n, m;
char s[MaxN + 5];
int succ[MaxN + 5];
bool vis[30];
int main() {
  while (~scanf("%d%d", &n, &m)) {
    scanf("%s", s);
    int len = strlen(s);
    memset(vis, 0, sizeof(vis));
    int Max, Min;
    for (int i = 0; i < len; i++) vis[s[i] - 'a'] = 1;
    for (int i = 0; i < 26; i++)
      if (vis[i]) {
        Min = i;
        break;
      }
    for (int i = 25; i >= 0; i--)
      if (vis[i]) {
        Max = i;
        break;
      }
    string res = "";
    if (m > n) {
      for (int i = 0; i < len; i++) res += s[i];
      for (int i = 1; i <= m - n; i++) res += char(Min + 'a');
    } else {
      succ[m] = 100;
      for (int i = m - 1; i >= 0; i--) succ[i] = min(succ[i + 1], s[i] - 'a');
      int pos = 1000000;
      for (int i = 0; i < n; i++) {
        if (Max <= succ[i + 1]) {
          pos = i;
          break;
        }
      }
      if (pos > m - 1) pos = m - 1;
      for (int i = 0; i < pos; i++) res += s[i];
      for (int i = s[pos] - 'a' + 1; i <= 25; i++)
        if (vis[i]) {
          res += char(i + 'a');
          break;
        }
      for (int i = 1; i <= m - (pos + 1); i++) res += char(Min + 'a');
    }
    cout << res << endl;
  }
  return 0;
}
