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;
void TxtRead() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
const int N = 1e5 + 5;
int n, k;
char s[N];
bool presnt[26];
vector<int> x;
map<char, int> mpx;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    presnt[s[i] - 'a'] = true;
  }
  for (int i = 0; i < 26; i++) {
    if (presnt[i]) {
      x.push_back(i);
      mpx[char('a' + i)] = x.size() - 1;
    }
  }
  if (k > n) {
    for (int i = 0; i < n; i++) cout << s[i];
    for (int i = 0; i < k - n; i++) cout << char('a' + x[0]);
  } else {
    int strt = 0, i = n - 1;
    while (mpx[s[i]] == x.size() - 1 || i > k - 1) {
      i--;
    }
    strt = i, i = 0;
    while (i != strt) cout << s[i], i++;
    cout << char('a' + x[mpx[s[strt]] + 1]);
    for (i = strt + 1; i < k; i++) cout << char(x[0] + 'a');
  }
}
