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;
long long a[2000010], d[2000010];
long long b[200010], f[1010][1010];
long long mn = 4e18, mx = -4e18, ans, sm = 1, cnt, n, m, cur;
bool ok, u[1000010], uu[310][310];
char c[510][510], cc[100010];
int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
vector<long long> g[900010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long k;
  cin >> n >> k;
  string s, t = "";
  cin >> s;
  vector<char> v;
  for (int i = 0; i < s.size(); i++) {
    ok = false;
    for (int j = 0; j < v.size(); j++) {
      if (v[j] == s[i]) {
        ok = true;
        break;
      }
    }
    if (!ok) v.push_back(s[i]);
  }
  sort(v.begin(), v.end());
  if (k > s.size()) {
    cout << s;
    for (int i = s.size(); i < k; i++) {
      cout << v[0];
    }
    return 0;
  } else if (k < s.size()) {
    t = s.substr(0, k);
    ok = false;
    for (int i = k - 1; i >= 0; i--) {
      for (int j = 0; j < v.size(); j++) {
        if (v[j] > t[i]) {
          t[i] = v[j];
          ok = true;
          break;
        }
      }
      if (ok) {
        for (int ii = i + 1; ii < k; ii++) {
          t[ii] = v[0];
        }
        cout << t;
        return 0;
      }
    }
  } else {
    t = s;
    ok = false;
    for (int i = k - 1; i >= 0; i--) {
      for (int j = 0; j < v.size(); j++) {
        if (v[j] > t[i]) {
          t[i] = v[j];
          ok = true;
          break;
        }
      }
      if (ok) {
        for (int ii = i + 1; ii < k; ii++) {
          t[ii] = v[0];
        }
        cout << t;
        return 0;
      }
    }
  }
  return 0;
}
