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 INF = 1e9 + 7;
const int MOD = 1e9 + 7;
int main() {
  int N, K;
  string S;
  cin >> N >> K;
  cin >> S;
  set<char> second;
  vector<char> V;
  for (int i = 0; i < (int)S.size(); i++) {
    second.insert(S[i]);
  }
  for (set<char>::iterator it = second.begin(); it != second.end(); it++) {
    V.push_back(*it);
  }
  if (N < K) {
    for (int i = 0; i < N; i++) {
      cout << S[i];
    }
    for (int i = 0; i < K - N; i++) {
      cout << V[0];
    }
    cout << endl;
  } else {
    vector<char> ans;
    for (int i = 0; i < K; i++) {
      ans.push_back(S[i]);
    }
    for (int i = K - 1; i >= 0; i--) {
      vector<char>::iterator it;
      it = upper_bound(V.begin(), V.end(), ans[i]);
      if (it == V.end()) {
        continue;
      } else {
        ans[i] = *it;
        for (int j = i + 1; j < K; j++) {
          ans[j] = V[0];
        }
        break;
      }
    }
    for (int i = 0; i < K; i++) {
      cout << ans[i];
    }
    cout << endl;
  }
  return 0;
}
