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 MOD = 1e9 + 7;
long long power(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    if (ret >= MOD) ret %= MOD;
    if (a >= MOD) a %= MOD;
    b >>= 1;
  }
  return ret;
}
long long inv(long long x) { return power(x, MOD - 2); }
const int N = 1e5 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, indx, a[26] = {0}, minm = 100, maxm = -1;
  string s, t = "";
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    a[s[i] - 'a'] = 1;
    maxm = max(maxm, s[i] - 'a');
    minm = min(minm, s[i] - 'a');
  }
  if (n >= k) {
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] - 'a' != maxm && i < k) {
        indx = i;
        break;
      }
    }
    for (int i = 0; i < indx; i++) {
      t += s[i];
    }
    for (int i = 0; i < 26; i++) {
      if (a[i] == 1 && i > s[indx] - 'a') {
        t += (i + 'a');
        break;
      }
    }
    for (int i = indx + 1; i < k; i++) t += (minm + 'a');
  } else {
    t += s;
    for (int i = 1; i <= k - n; i++) t += (minm + 'a');
  }
  cout << t << endl;
}
