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 myRnd() { return abs(((rand() << 15) ^ rand())); }
int myRnd(int L, int R) {
  return abs(((rand() << 15) ^ rand())) % (R - L + 1) + L;
}
void Parr(int *arr, int L, int R) {
  for (int i = L; R >= i; i++) {
    printf("%d%c", arr[i], " \n"[i == R]);
  }
}
void Pvec(vector<int> v) {
  for (int i = 0; ((int)(v).size()) > i; i++) {
    printf("%d%c", v[i], " \n"[i == ((int)(v).size()) - 1]);
  }
}
void Sarr(int *arr, int L, int R) {
  for (int i = L; R >= i; i++) {
    scanf("%d", &(arr[i]));
    ;
  }
}
const int N = 2e5 + 6;
int a[N];
int main() {
  srand(time(NULL));
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  for (char i : s) a[i] = 1;
  if (k > n) {
    cout << s;
    int delta = (k - n);
    for (int i = 'a'; 'z' >= i; i++) {
      if (a[i]) {
        while (delta--) cout << char(i);
        break;
      }
    }
    cout << endl;
    return 0;
  }
  string ans;
  for (int i = 0; k > i; i++) ans.push_back(s[i]);
  for (int i = k - 1; i >= 0; i--) {
    bool get_ans = false;
    for (int x = ans[i] + 1; 'z' >= x; x++) {
      if (a[x]) {
        ans[i] = x;
        get_ans = true;
        break;
      }
    }
    if (get_ans) {
      cout << ans << endl;
      return 0;
    }
    for (int x = 'a'; 'z' >= x; x++) {
      if (a[x]) {
        ans[i] = x;
        break;
      }
    }
  }
}
