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 t[100010];
bool alp[150];
char s, e;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  cin >> t;
  for (int i = 0; t[i]; i++) alp[t[i]] = true;
  for (int i = 'a'; i <= 'z'; i++) {
    if (alp[i]) {
      s = i;
      break;
    }
  }
  for (int i = 'z'; i >= 'a'; i--) {
    if (alp[i]) {
      e = i;
      break;
    }
  }
  if (k > n) {
    for (int l = n; l < k; l++) t[l] = s;
    for (int l = 0; l < k; l++) cout << t[l];
    return 0;
  } else {
    for (int i = k - 1; i > -1; i--) {
      if (t[i] < e) {
        for (int j = t[i] + 1; j <= 'z'; j++) {
          if (alp[j]) {
            t[i] = j;
            for (int l = i + 1; l < k; l++) {
              t[l] = s;
            }
            for (int l = 0; l < k; l++) cout << t[l];
            return 0;
          }
        }
      }
    }
  }
  return 0;
}
