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 main() {
  int n, k;
  char node[100005];
  char t[100005];
  char min = 'z';
  char max = 'a';
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> node[i];
    t[i] = node[i];
    if (min > node[i]) min = node[i];
    if (max < node[i]) max = node[i];
  }
  sort(t, t + n);
  if (k > n) {
    for (int i = 0; i < n; i++) {
      cout << node[i];
    }
    for (int j = 0; j < k - n; j++) {
      cout << min;
    }
    cout << endl;
  } else {
    int i;
    for (i = k - 1; i >= 0; i--) {
      if (node[i] != max) {
        break;
      }
    }
    char x;
    for (int j = 0; j < n; j++) {
      if (t[j] > node[i]) {
        x = t[j];
        break;
      }
    }
    for (int p = 0; p < i; p++) {
      cout << node[p];
    }
    cout << x;
    for (int q = i + 1; q < k; q++) {
      cout << min;
    }
    cout << endl;
  }
  return 0;
}
