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 n, k;
bool vis[27];
string s;
vector<int> vec;
vector<int> a, b;
int pos[27], rpos[27];
int main(int argc, char *argv[]) {
  memset(vis, 0, sizeof(vis));
  char ch;
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> ch;
    a.push_back(ch - 'a' + 1);
    if (vis[ch - 'a' + 1])
      ;
    else
      vis[ch - 'a' + 1] = 1, vec.push_back(ch - 'a' + 1);
  }
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) pos[vec[i]] = i, rpos[i] = vec[i];
  for (int i = 0; i < n; i++) b.push_back(a[i]), a[i] = pos[a[i]];
  if (n == k) {
    a[n - 1] += 1;
    for (int i = n - 1; i >= 1; i--)
      if (a[i] >= vec.size()) {
        a[i] -= vec.size();
        a[i - 1]++;
      }
    for (int i = 0; i < n; i++) cout << char(rpos[a[i]] + 'a' - 1);
  } else if (n < k) {
    while (b.size() != k) b.push_back(vec[0]);
    for (int i = 0; i < b.size(); i++) cout << char(b[i] + 'a' - 1);
  } else {
    a[k - 1] += 1;
    for (int i = k - 1; i >= 1; i--)
      if (a[i] >= vec.size()) {
        a[i] -= vec.size();
        a[i - 1]++;
      }
    for (int i = 0; i < k; i++) cout << char(rpos[a[i]] + 'a' - 1);
  }
  return 0;
}
