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 cmmdc(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lgput(long long cnt, long long mod, long long put) {
  long long r = 1, p = put;
  if (cnt == 0)
    return 1;
  else if (cnt < 0)
    return 0;
  while (cnt > 1) {
    if (!(cnt & 1)) {
      p = p * 1LL * p % mod;
      cnt >>= 1LL;
    } else {
      --cnt;
      r = r * 1LL * p % mod;
    }
  }
  p = p * 1LL * r;
  return p % mod;
}
int dist(int a, int b) {
  if (a > b) return a - b;
  return b - a;
}
char v[100005], fv[500];
int main() {
  long long a, b, nrca, nrcb, n, i, j, c, ca, cb, mx = -9999999999,
                                                  mn = 9999999999, s = 0, k;
  char val = 'z' + 1, vlm = 'a' - 1;
  cin >> n >> k >> v;
  for (i = 0; i < n; ++i) {
    ++fv[v[i]];
    if (v[i] < val) val = v[i];
    if (v[i] > vlm) vlm = v[i];
  }
  if (n < k) {
    for (i = 0; i < n; ++i) cout << v[i];
    for (i = 1; i <= k - n; ++i) cout << val;
    return 0;
  }
  for (i = k - 1; i > 0; --i)
    if (v[i] != vlm) break;
  for (j = v[i] + 1; j <= 'z'; ++j)
    if (fv[j]) break;
  v[i] = j;
  for (++i; i < k; ++i) v[i] = val;
  for (i = 0; i < k; ++i) cout << v[i];
  return 0;
}
