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>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define FORR(i, n) for(ll i = (ll)n - 1LL; i >= 0LL; i--)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
using ll = long long;
template <typename T> using V = vector<T>;
constexpr int Mod = 998244353;
constexpr int mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;
template <typename T> constexpr bool chmax(T &a, const T &b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> constexpr bool chmin(T &a, const T &b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

int N, K;
string S;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    cin >> N >> K >> S;
    string t = S;
    S += string(S.rbegin(), S.rend());
    FOR(i, 1, N + 1) chmin(t, S.substr(i, N));
    int l = 1;
    while(l < t.size() && t[0] == t[l])
        l++;
    int k = l;
    if(--K > 15)
        k = N;
    else
        k <<= K;
    string ans;
    rep(i, min(k, N)) ans.push_back(t[0]);
    ans += t.substr(l, N - ans.size());
    cout << ans << endl;

    return 0;
}