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>
#pragma GCC optimize(2)
using namespace std;
int n, k, small = 0, last = 0, vis[205], s[105000];
char p;
int main(int argc, char const *argv[]) {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    cin >> p;
    vis[p - 'a'] = true;
    s[i] = p - 'a';
  }
  for (int i = 0; i <= 26; i++)
    if (vis[i]) {
      small = i;
      break;
    }
  for (int i = 0; i <= 26; i++)
    if (vis[i]) last = i;
  if (k > n) {
    for (int i = 1; i <= n; i++) cout << (char)(s[i] + 'a');
    for (int i = 1; i <= (k - n); i++) cout << (char)(small + 'a');
  } else if (k == n) {
    int loc = n;
    for (int i = n; i >= 1; i--)
      if (s[i] != last) {
        loc = i;
        break;
      }
    for (int i = 1; i < loc; i++) cout << (char)(s[i] + 'a');
    int nex = 0;
    for (int i = s[loc] + 1; i <= 26; i++)
      if (vis[i]) {
        nex = i;
        break;
      }
    cout << (char)('a' + nex);
    for (int i = loc + 1; i <= n; i++) cout << (char)(small + 'a');
  } else {
    int loc = k;
    for (int i = k; i >= 1; i--)
      if (s[i] != last) {
        loc = i;
        break;
      }
    for (int i = 1; i < loc; i++) cout << (char)(s[i] + 'a');
    int nex = 0;
    for (int i = s[loc] + 1; i <= 26; i++)
      if (vis[i]) {
        nex = i;
        break;
      }
    cout << (char)('a' + nex);
    for (int i = loc + 1; i <= k; i++) cout << (char)(small + 'a');
  }
  return 0;
}
