Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:

* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.



Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.

Constraints

* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


N K
S


Output

Print the lexicographically smallest possible string that can be the string S after the K operations.

Examples

Input

5 1
bacba


Output

aabca


Input

10 2
bbaabbbaab


Output

aaaabbaabb
#include <bits/stdc++.h>

using namespace std;


void solve() {
    int n, k;
    cin >> n >> k;
    string s; cin >> s;
    string t = s; reverse(t.begin(), t.end());
    string u = s + t;
    char mi = 'z';
    for (char x: s) {
        mi = min(mi, x);
    }
    int mx = 1;
    vector<int> len(2*n+1, 0);
    for (int i = 0; i < 2*n; i++) {
        len[i+1] = u[i] == mi ? len[i]+1 : 0;
        len[i+1] = min(len[i+1], n);
        mx = max(mx, len[i+1]);
    }
    if (k>15 || (1<<(k-1))*mx >= n) {
        for (int i = 0; i < n; i++) {
            cout << mi;
        }
        return;
    }
    int rep =(1<<(k-1)) * mx;
    int m = n - rep;
    string res = "{";
    for (int i = n+1; i <= 2*n; i++) {
        if (len[i] == mx) {
            string tmp = u.substr(i-m-mx, m);
            reverse(tmp.begin(), tmp.end());
            res = min(res, tmp);
        }
    }
    for (int i = 0; i < rep; i++) {
        cout << mi;
    }
    cout << res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}
