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;
char s[100100], ans[100100];
bool ch[100];
bool out[1011000];
int ex[30];
int id;
int main() {
  cin >> n >> k;
  scanf("%s", &s);
  for (int i = 0; i < n; i++) {
    ch[s[i] - 'a'] = true;
  }
  int maxx = 0;
  for (int i = 0; i < 27; i++) {
    if (ch[i]) {
      ex[id++] = i;
      maxx = i;
    }
  }
  if (k <= n) {
    for (int i = k - 1; i >= 0; i--) {
      if ((int)(s[i] - 'a') == maxx) {
        out[i] = true;
        ans[i] = ex[0] + 'a';
      } else {
        for (int j = 0; j < id; j++) {
          if ((int)s[i] < ex[j] + 'a') {
            out[i] = true;
            ans[i] = (char)(ex[j] + 'a');
            break;
          }
        }
        break;
      }
    }
    string st = "";
    for (int i = 0; i < k; i++) {
      if (!out[i])
        st += s[i];
      else
        st += ans[i];
    }
    cout << st;
    cout << "\n";
  } else {
    string st = "";
    st += s;
    for (int i = n; i < k; i++) {
      st += (ex[0] + 'a');
    }
    cout << st << endl;
  }
  return 0;
}
