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>
#pragma comment(linker, "/stack:156000000")
using namespace std;
const int MOD = (int)1e9 + 7;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int MAXN = (int)1e5 + 5;
int n, m, k, x, y, t;
int a[MAXN], b[MAXN], c[MAXN];
string s, w;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < min(n, k); i++) w += s[i];
  vector<char> c(0);
  for (int i = 0; i < n; i++) c.push_back(s[i]);
  sort(c.begin(), c.end());
  c.erase(unique(c.begin(), c.end()), c.end());
  if (k > n) {
    for (int i = w.size(); i < k; i++) w += c[0];
    cout << w;
    return 0;
  }
  for (int i = k - 1; i >= 0; i--) {
    if (w[i] == c.back()) continue;
    for (int j = c.size() - 1; j >= 0; j--)
      if (w[i] == c[j]) {
        w[i] = c[j + 1];
        for (int t = i + 1; t < k; t++) w[t] = c[0];
        cout << w;
        return 0;
      }
  }
  cout << "huy";
}
