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;
const int maxn = (int)1e5 + 9;
const long long mod = 987654323;
char str[maxn];
int mn, mx;
int n, k;
int vis[300];
void init() {
  scanf("%d%d%s", &n, &k, str);
  mn = 10000;
  mx = -1000;
  for (int i = 0; i < n; i++) {
    vis[str[i]] = 1;
    mn = min(mn, (int)str[i]);
    mx = max(mx, (int)str[i]);
  }
}
int main() {
  init();
  if (n < k) {
    printf("%s", str);
    for (int i = n; i < k; i++) {
      printf("%c", mn);
    }
    return 0;
  }
  int tmp = k - 1;
  while (str[tmp] == mx) tmp--;
  for (int i = 0; i < tmp; i++) {
    printf("%c", str[i]);
  }
  for (int i = str[tmp] + 1; i < 300; i++) {
    if (vis[i]) {
      printf("%c", i);
      break;
    }
  }
  for (int i = tmp + 1; i < k; i++) {
    printf("%c", mn);
  }
  return 0;
}
