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 int N = (int)2e5 + 7;
const int MOD = (int)1e9 + 9;
using namespace std;
char second[N];
bool ok[26];
int main() {
  long long n, t;
  cin >> n >> t;
  for (int i = 1; i <= n; i++) {
    cin >> second[i];
    ok[second[i] - 'a'] = 1;
  }
  if (t > n) {
    for (int i = 1; i <= n; i++) {
      cout << second[i];
    }
    for (int i = n + 1; i <= t; i++) {
      for (int j = 0; j < 26; j++) {
        if (ok[j]) {
          cout << char('a' + j);
          break;
        }
      }
    }
    return 0;
  }
  long long ans = -1;
  for (int i = n; i >= 1; i--) {
    long long c = second[i] - 'a';
    if (i > t) continue;
    for (int j = c + 1; j < 26; j++) {
      if (ok[j] == true) {
        ans = i;
        break;
      }
    }
    if (ans != -1) break;
  }
  bool ok1 = false;
  for (int i = 1; i <= t; i++) {
    if (ok1) {
      for (int j = 0; j < 26; j++) {
        if (ok[j]) {
          cout << char('a' + j);
          break;
        }
      }
      continue;
    }
    long long c = second[i] - 'a';
    if (i == ans) {
      ok1 = true;
      for (int j = c + 1; j < 26; j++) {
        if (ok[j]) {
          cout << char('a' + j);
          break;
        }
      }
      continue;
    }
    cout << second[i];
  }
  return 0;
}
