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;
void files(bool in, char *s) {
  if (in)
    freopen(s, "r", stdin);
  else
    freopen(s, "w", stdout);
}
const double PI = 4 * atan(1.0);
const double EPS = 1e-9;
const int infi = (int)1e9;
const long long infl = (long long)1e17;
const int MOD = (int)1e9 + 7;
const int MAX = (int)1e6 + 22;
const int dx[] = {0, 1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 0, -1, 1, 1, -1, 1, -1};
const int kdx[] = {1, 1, -1, -1, 2, 2, -2, -2};
const int kdy[] = {2, -2, 2, -2, 1, -1, 1, -1};
long long c[1234];
long long rose[123];
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int mx = 0;
  int mn = 1e9;
  int x;
  for (int i = 0; i < n; i++) {
    x = s[i];
    c[x] = 1;
    mx = max(x, mx);
    mn = min(x, mn);
  }
  if (k > n) {
    for (int i = n; i < k; i++) {
      s += (char)mn;
    }
    cout << s << "\n";
    return 0;
  }
  for (int i = k - 1; i >= 0; i--) {
    if (s[i] == mx) {
      s[i] = (char)mn;
    } else {
      for (int j = (int)s[i] + 1; j <= (int)mx; j++) {
        if (c[j]) {
          s[i] = (char)j;
          for (int h = 0; h < k; h++) cout << s[h];
          return 0;
        }
      }
    }
  }
  return 0;
}
