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;
set<string> ans;
set<char> lt;
set<pair<int, int> > hashing;
int nums[100005];
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, k, l, c = 0, n, m, total = 0, a, b, x, y;
  string s, t;
  cin >> n >> k;
  cin >> s;
  nums[0] = 0;
  for (i = 1; i < n + 1; i = i + 1)
    nums[i] += nums[i - 1] + s[i - 1], lt.insert(s[i - 1]);
  set<char>::iterator it = lt.begin();
  a = int(*it);
  b = 0;
  if (k > n) {
    t = s;
    for (i = n; i < k; i = i + 1) {
      t = t + char(a);
    }
    cout << t;
    return 0;
  }
  for (i = 1; i < k + 1; i = i + 1) {
    b = nums[i - 1];
    it = lt.find(s[i - 1]);
    it++;
    if (it != lt.end()) {
      b += int(*it) + (k + 1 - i) * a;
      hashing.insert(make_pair(i - 1, b));
    } else {
      hashing.insert(make_pair(i - 1, 1e9 + 7));
    }
  }
  set<pair<int, int> >::iterator gh = hashing.end();
  gh--;
  y = 1e9 + 7;
  for (; gh != hashing.begin(); gh--) {
    x = (*gh).first;
    y = (*gh).second;
    if (y != 1e9 + 7) {
      break;
    }
  }
  gh = hashing.begin();
  if (y == 1e9 + 7) x = (*gh).first;
  t = s;
  it = lt.find(s[x]);
  it++;
  t[x] = *it;
  for (i = x + 1; i < n; i = i + 1) t[i] = char(a);
  for (i = 0; i < k; i = i + 1) cout << t[i];
  return 0;
}
