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;
char a[100011];
void solve() {
  int n, k;
  scanf("%d %d", &n, &k);
  scanf("%s", a);
  set<char> s;
  for (int i = 0; i < n; i++) s.insert(a[i]);
  if (k > n) {
    printf("%s", a);
    char c = *(s.begin());
    for (int i = n; i < k; i++) printf("%c", c);
    printf("\n");
    return;
  }
  set<char>::iterator it = s.end();
  it--;
  char c = *(it);
  int v = k - 1;
  while (a[v] == c) v--;
  a[v] = *(s.upper_bound(a[v]));
  c = *(s.begin());
  for (int i = v + 1; i < k; i++) a[i] = c;
  for (int i = 0; i < k; i++) printf("%c", a[i]);
  printf("\n");
}
int main() {
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
