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 maxn = 1e5 + 5;
const int mod = 1000000007;
char a[maxn], b[maxn];
char mm = 127;
bool vis[256];
int n, k;
char find(char x) {
  while (++x <= 'z')
    if (vis[x]) return x;
  return 'z' + 1;
}
int main() {
  cin >> n >> k;
  cin >> a + 1;
  memcpy(b + 1, a + 1, k);
  for (int i = 1; i <= n; ++i) vis[a[i]] = true, mm = min(mm, a[i]);
  if (k > n) {
    for (int i = n + 1; i <= k; ++i) b[i] = mm;
    cout << b + 1 << endl;
    return 0;
  }
  for (int i = k; i >= 1; --i) {
    char c = find(b[i]);
    if (c <= 'z')
      b[i] = c;
    else
      continue;
    for (int j = i + 1; j <= k; ++j) b[j] = mm;
    cout << b + 1 << endl;
    return 0;
  }
  return 0;
}
