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 long long INF = 1e18;
long long my_pow(long long a, long long n) {
  int res = 1;
  while (n)
    if (n & 1) {
      res *= a;
      --n;
    } else {
      a *= a;
      n >>= 1;
    }
  return res;
}
bool a[32];
int main() {
  int n, k;
  string s, t;
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < n; i++) a[(int)s[i] - 96] = true;
  char c;
  for (int i = 1; i < 27; i++)
    if (a[i]) {
      c = (char)(i + 96);
      break;
    }
  for (int i = 0; i < k; i++) t += (i < n ? s[i] : c);
  if (k > n) {
    cout << t;
    return 0;
  }
  int e = 0;
  for (int i = k - 1; i >= 0; i--) {
    for (int j = 1; j <= 26; j++) {
      if (a[j] && j > (int)s[i] - 96) {
        t[i] = (char)(j + 96);
        e = i + 1;
        break;
      }
    }
    if (e > 0) break;
  }
  for (int i = e; i < k; i++) t[i] = c;
  cout << t;
  return 0;
}
