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;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = ((long long)1 << 61) - 1;
int len;
bool flag[27] = {0};
int main() {
  int n, k;
  string a, b;
  cin >> n >> k;
  cin >> a;
  for (int i = 0; i < a.length(); i++) {
    flag[a[i] - 'a'] = true;
  }
  if (k > n) {
    cout << a;
    char c = 0;
    for (int j = 0; j < 26; j++) {
      if (flag[j]) {
        c = 'a' + j;
        break;
      }
    }
    for (int i = a.length(); i < k; i++) cout << c;
    cout << endl;
  } else {
    int f = 0;
    char c = 0;
    for (int j = 0; j < 26; j++) {
      if (flag[j]) {
        c = 'a' + j;
        break;
      }
    }
    for (int i = k - 1; i >= 0; i--) {
      if (f) {
        b += a[i];
        continue;
      }
      for (int j = a[i] - 'a' + 1; j < 26; j++) {
        if (flag[j]) {
          b += 'a' + j;
          f = 1;
          break;
        }
      }
      if (!f) b += c;
    }
    for (int i = b.length() - 1; i >= 0; i--) cout << b[i];
    cout << endl;
  }
  return 0;
}
