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 mymin(vector<long long int> v) { return *min_element(v.begin(), v.end()); }
long long int mymax(vector<long long int> v) {
  return *max_element(v.begin(), v.end());
}
long long int power(long long int a, long long int b, long long int p) {
  long long int res = 1;
  a %= p;
  while (b != 0) {
    if (b % 2) res = (res % p * a % p) % p;
    b /= 2;
    a = (a % p * a % p) % p;
  }
  return res;
}
vector<string> split(string str, char delimeter) {
  vector<string> res;
  istringstream iss(str);
  string token;
  while (getline(iss, token, delimeter)) {
    if (token.length() > 0) res.push_back(token);
  }
  return res;
}
void randomize(vector<int> &arr) {
  srand(time(NULL));
  for (int i = arr.size() - 1; i > 0; i--) {
    int j = rand() % (i + 1);
    swap(arr[i], arr[j]);
  }
}
long long int mgcd(long long int a, long long int b) {
  if (a == 0) return b;
  return mgcd(b % a, a);
}
int main() {
  clock_t tstart = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  string s;
  cin >> n >> k >> s;
  set<char> st;
  for (int i = 0; i < n; i++) st.insert(s[i]);
  char bg = *st.begin();
  char last = bg;
  for (auto x : st) {
    last = x;
  }
  if (k > n) {
    cout << s;
    for (int i = 0; i < k - n; i++) cout << bg;
    return 0;
  }
  char ans[k];
  for (int i = 0; i < k; i++) ans[i] = s[i];
  int ind = k - 1;
  while (ans[ind] == last) ind--;
  if (ind != k - 1) {
    char ch1 = ans[ind];
    set<char>::iterator it1 = st.find(ch1);
    it1++;
    ans[ind] = *it1;
    for (int i = ind + 1; i < k; i++) ans[i] = bg;
    for (int i = 0; i < k; i++) cout << ans[i];
    return 0;
  }
  for (int i = 0; i < k - 1; i++) cout << s[i];
  char ch = s[k - 1];
  set<char>::iterator it = st.find(ch);
  it++;
  cout << *it;
  return 0;
}
