Identify the time complexity of the following code as constant, linear, quadratic, cubic, log(n), nlog(n) or NP-hard. import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Iterator;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.NoSuchElementException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Egor Kulikov (egor@egork.net)
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        TaskE2 solver = new TaskE2();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++) {
            solver.solve(i, in, out);
        }
        out.close();
    }

    static class TaskE2 {
        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int n = in.readInt();
            int m = in.readInt();
            int[][] a = in.readIntTable(n, m);
            int[][] id = new int[n + 1][1 << n];
            int[][] val = new int[n + 1][1 << n];
            ArrayUtils.fill(id, m);
            int[] sum = new int[1 << n];
            int[] low = new int[1 << n];
            boolean[] base = new boolean[1 << n];
            int[] vv = new int[1 << n];
            for (int i = 1; i < (1 << n); i++) {
                low[i] = Integer.bitCount(Integer.lowestOneBit(i) - 1);
                int current = i;
                base[i] = true;
                vv[i] = i;
                for (int j = 0; j < n; j++) {
                    current = current << 1;
                    current += current >> n;
                    current -= (current >> n) << n;
                    if (current < i) {
                        base[i] = false;
              ... Complexity: