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;
bool a[1007];
unordered_map<char, char> mx;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m, i, j;
  string second, ans = "";
  char lowest, highest;
  cin >> n >> m >> second;
  for (i = 0; i < n; i++) a[second[i]] = 1;
  for (i = 'a'; i <= 'z'; i++) {
    bool x = 0;
    for (j = i + 1; j <= 'z'; j++) {
      if (a[(char)j]) {
        mx[(char)i] = (char)j;
        x = 1;
        break;
      }
    }
    if (!x) {
      for (j = 'a'; j < i; j++)
        if (a[(char)j]) {
          mx[(char)i] = (char)j;
          x = 1;
          break;
        }
    }
  }
  for (i = 'a'; i <= 'z'; i++)
    if (a[i]) {
      lowest = char(i);
      break;
    }
  for (i = 'z'; i >= 'a'; i--)
    if (a[i]) {
      highest = char(i);
      break;
    }
  if (m > n) {
    ans = second;
    for (i = n; i < m; i++) ans += lowest;
  } else if (m == n) {
    for (i = n - 1; i >= 0; i--)
      if (mx[second[i]] > second[i]) {
        j = i;
        break;
      }
    for (i = 0; i < j; i++) ans += second[i];
    for (i = j; i < n; i++) ans += mx[second[i]];
  } else if (m < n) {
    for (i = m - 1; i >= 0; i--)
      if (mx[second[i]] > second[i]) {
        j = i;
        break;
      }
    for (i = 0; i < j; i++) ans += second[i];
    for (i = j; i < m; i++) ans += mx[second[i]];
  }
  cout << ans << endl;
  return 0;
}
