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;
set<char> st;
string s, t;
int n, k, dp[100005][2];
int solve(int i, int f) {
  if (i == k) return f;
  if (dp[i][f] != -1) return dp[i][f];
  int ans = 0;
  for (char c : st) {
    if (!f && c < s[i]) continue;
    ans = max(ans, solve(i + 1, f ? 1 : (c > s[i])));
    if (ans) {
      t[i] = c;
      break;
    }
  }
  return dp[i][f] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  while (cin >> n >> k) {
    cin >> s;
    st.clear();
    for (char c : s) st.insert(c);
    if (k > n) {
      cout << s;
      for (int i = 0; i < k - n; i++) cout << *st.begin();
      cout << '\n';
    } else {
      memset(dp, -1, sizeof dp);
      t = s;
      solve(0, 0);
      cout << t.substr(0, k) << '\n';
    }
  }
  return 0;
}
