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;
vector<int> v;
map<char, long long int> mp;
const long long int INF = 1e18;
long long int cnt;
int main() {
  long long int n, x = 0, y, m, f, d, p = 0, k, a, b, i = 0, j;
  string s, r = "";
  cin >> n >> m;
  cin >> s;
  for (i = 0; i < s.length(); i++) {
    mp[s[i]] = 1;
  }
  if (n < m) {
    r = s;
    sort(s.begin(), s.end());
    for (i = r.length(); i < m; i++) r += s[0];
    return cout << r, 0;
  }
  for (i = m - 1; i >= 0; i--) {
    char c = s[i], d;
    if (c == 'z')
      c = 'a';
    else
      c++;
    while (mp[c] == 0) {
      if (c == 'z')
        c = 'a';
      else {
        c++;
      }
    }
    if (c < s[i]) {
      x = 0;
      r += c;
    } else {
      x = 1;
      r += c;
      for (k = i - 1; k >= 0; k--) {
        r += s[k];
      }
      reverse(r.begin(), r.end());
      return cout << r, 0;
    }
  }
}
