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, len, cnt;
char s[100010];
int vis[300], a[100010];
int d[300];
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  len = strlen(s + 1);
  memset(vis, -1, sizeof(vis));
  for (int i = 1; i <= len; i++)
    if (vis[s[i]] == -1) vis[s[i]] = 1;
  for (int i = 'a'; i <= 'z'; i++)
    if (vis[i] == 1) vis[i] = cnt, d[cnt++] = i;
  for (int i = 1; i <= len; i++) a[i] = vis[s[i]];
  if (k > len) {
    printf("%s", s + 1);
    for (int i = len + 1; i <= k; i++) printf("%c", d[0]);
    return 0;
  }
  a[k]++;
  int nw = k;
  while (k) {
    a[k - 1] += a[k] / cnt;
    a[k] %= cnt;
    k--;
  }
  for (int i = 1; i <= nw; i++) printf("%c", d[a[i]]);
}
