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;
int n, kk, k, used[1000000], c, thien, tmp;
char tt, a[1000000], s[1000001];
int main() {
  cin >> n >> kk;
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 1; i <= n; i++)
    if (used[s[i]] == 0) {
      used[s[i]] = 1;
    }
  for (char i = 'a'; i <= 'z'; i++) {
    if (used[i] != 0) {
      c++;
      a[c] = i;
      if (c == 1) tt = i;
    }
  }
  thien = -1;
  if (n < kk) {
    for (int i = 1; i <= kk; i++)
      if (i <= n)
        cout << s[i];
      else
        cout << tt;
  } else {
    k = kk;
    for (int i = k; i >= 1; i--) {
      for (int j = 1; j <= c; j++) {
        if (a[j] - s[i] > 0) {
          thien = i;
          tmp = j;
          break;
        }
      }
      if (thien != -1) break;
    }
    for (int i = 1; i < thien; i++) cout << s[i];
    cout << a[tmp];
    for (int i = thien + 1; i <= k; i++) cout << tt;
  }
}
