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 int inf = 1e18;
const long long int mod = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long int t, n, k, p, i, s, c, ind;
  cin >> n >> k;
  string str;
  cin >> str;
  char ch, maxi, res;
  set<char> st;
  vector<char> v;
  st.insert(str[0]), v.push_back(str[0]);
  maxi = str[0];
  for (long long int i = 1; i <= n - 1; i++) {
    ch = str[i];
    if (st.find(ch) == st.end()) {
      st.insert(ch);
      v.push_back(ch);
      if (str[i] > maxi) maxi = str[i];
    }
  }
  sort(v.begin(), v.end());
  s = v.size();
  if (k <= n) {
    i = k - 1;
    while (str[i] == maxi) i--;
    ch = str[i];
    ind = i;
    for (long long int j = 0; j < s; j++) {
      if (v[j] > ch) {
        res = v[j];
        break;
      }
    }
    for (long long int i = 0; i < k; i++) {
      if (i < ind)
        cout << str[i];
      else if (i == ind)
        cout << res;
      else
        cout << v[0];
    }
    cout << endl;
  } else {
    cout << str;
    for (long long int i = 0; i < k - n; i++) cout << v[0];
    cout << endl;
  }
}
