A two dimensional array is called a bracket array if each grid contains one of the two possible brackets — "(" or ")". A path through the two dimensional array cells is called monotonous if any two consecutive cells in the path are side-adjacent and each cell of the path is located below or to the right from the previous one. 

A two dimensional array whose size equals n × m is called a correct bracket array, if any string formed by writing out the brackets on some monotonous way from cell (1, 1) to cell (n, m) forms a correct bracket sequence. 

Let's define the operation of comparing two correct bracket arrays of equal size (a and b) like that. Let's consider a given two dimensional array of priorities (c) — a two dimensional array of same size, containing different integers from 1 to nm. Let's find such position (i, j) in the two dimensional array, that ai, j ≠ bi, j. If there are several such positions, let's choose the one where number ci, j is minimum. If ai, j = "(", then a < b, otherwise a > b. If the position (i, j) is not found, then the arrays are considered equal.

Your task is to find a k-th two dimensional correct bracket array. It is guaranteed that for the given sizes of n and m there will be no less than k two dimensional correct bracket arrays.

Input

The first line contains integers n, m and k — the sizes of the array and the number of the sought correct bracket array (1 ≤ n, m ≤ 100, 1 ≤ k ≤ 1018). Then an array of priorities is given, n lines each containing m numbers, number pi, j shows the priority of character j in line i (1 ≤ pi, j ≤ nm, all pi, j are different).

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Output

Print the k-th two dimensional correct bracket array.

Examples

Input

1 2 1
1 2


Output

()


Input

2 3 1
1 2 3
4 5 6


Output

(()
())


Input

3 2 2
3 6
1 4
2 5


Output

()
)(
()

Note

In the first sample exists only one correct two-dimensional bracket array.

In the second and in the third samples two arrays exist.

A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.Comparator;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.math.BigInteger;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author AlexFetisov
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskC solver = new TaskC();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskC {
    int N;
    private long k;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int m = in.nextInt();
        k = in.nextLong();
        int[][] initP = new int[n][m];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                initP[i][j] = in.nextInt();
            }
        }

        N = n + m - 1;
        final int[] p = new int[N];
        Arrays.fill(p, Integer.MAX_VALUE);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                p[i + j] = Math.min(p[i + j], initP[i][j]);
            }
        }
        final Integer[] idx = new Integer[N];
        for (int i = 0; i < N; ++i) {
            idx[i] = i;
        }
        Arrays.sort(idx, new Comparator<Integer>() {
            public int compare(Integer o1, Integer o2) {
                return p[o1] - p[o2];
            }
        });

        char[] brackets = new char[N];
        Arrays.fill(brackets, ' ');
        for (int i = 0; i < N; ++i) {
            int pos = idx[i];
            brackets[pos] = '(';
            long ways = getDp(brackets);
            if (ways < k) {
                brackets[pos] = ')';
                k -= ways;
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                out.print(brackets[i + j]);
            }
            out.println();
        }
    }

    private long getDp(char[] brackets) {
        long[][] dp = new long[N + 1][N + 1];
        dp[0][0] = 1;
        for (int i = 0; i < N; ++i) {
            for (int bal = 0; bal < N; ++bal) {
                if (dp[i][bal] > k) {
                    dp[i][bal] = k;
                }
                if (brackets[i] != '(' && bal > 0) {
                    dp[i+1][bal-1] += dp[i][bal];
                }
                if (brackets[i] != ')') {
                    dp[i+1][bal+1] += dp[i][bal];
                }
            }
        }
        return dp[N][0];
    }
}

class InputReader {
    private BufferedReader reader;
    private StringTokenizer stt;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
    }

    public String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            return null;
        }
    }

    public String nextString() {
        while (stt == null || !stt.hasMoreTokens()) {
            stt = new StringTokenizer(nextLine());
        }
        return stt.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(nextString());
    }

    public long nextLong() {
        return Long.parseLong(nextString());
    }

}

