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 INF = 2e9 + 999;
const long long LINF = 2e18 + 999;
void inv(bool e);
template <typename T>
T sign(T a);
template <typename T>
void die(T &a);
bool brute, alter;
int cnt_tests = 1;
string s;
int k;
void inp() { cin >> k >> k >> s; }
void solve() {
  set<char> order;
  for (auto c : s) {
    order.insert(c);
  }
  string t = s;
  if (k > ((int)(s).size())) {
    while (((int)(t).size()) != k) {
      t.push_back(*order.begin());
    }
    cout << t << '\n';
    return;
  }
  t.resize(k);
  bool need = true;
  for (int i = k - 1; i >= 0; --i) {
    if (need) {
      if (order.upper_bound(t[i]) != order.end()) {
        t[i] = (*order.upper_bound(t[i]));
        need = false;
      } else {
        t[i] = (*order.begin());
      }
    }
  }
  cout << t << '\n';
}
void stress() {}
void naive() {}
void run();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (0) {
    freopen("movetofront.in", "r", stdin);
    freopen("movetofront.out", "w", stdout);
  }
  srand(time(0));
  brute = false;
  for (int i = 0; (i < cnt_tests); i++) {
    run();
  }
  return 0;
}
void run() {
  if (!brute) {
    inp();
  } else {
    stress();
  }
  solve();
  cout << '\n';
}
template <typename T>
void die(T &a) {
  cout << a;
  exit(0);
}
template <typename T>
T sign(T a) {
  if (a == 0) return 0;
  if (a < 0) return -1;
  return 1;
}
template <typename T>
void umax(T &a, T b) {
  if (b > a) a = b;
}
template <typename T>
void umin(T &a, T b) {
  if (b < a) a = b;
}
void inv(bool e) {
  if (!e) {
    vector<int> a;
    a[-1] += 1;
  }
}
