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 static java.lang.Math.*;
import static java.lang.System.currentTimeMillis;
import static java.lang.System.exit;
import static java.lang.System.arraycopy;
import static java.util.Arrays.sort;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.fill;
import java.util.*;
import java.io.*;

public class Main {

	public static void main(String[] args) throws IOException {
		try {
			if (new File("input.txt").exists())
				System.setIn(new FileInputStream("input.txt"));
		} catch (SecurityException e) {
		}
		new Thread(null, new Runnable() {
			public void run() {
				try {
					new Main().run();
				} catch (Throwable e) {
					e.printStackTrace();
					exit(999);
				}
			}
		}, "1", 1 << 23).start();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st = new StringTokenizer("");

	int n, m;
	int len;
	int pr[] = new int[200];
	int inf = Integer.MAX_VALUE;
	
	int val[] = new int[200];
	boolean set[] = new boolean[200];
	long INF = Long.MAX_VALUE;
	long get(){
		long dp[][] = new long[200][200];
		if(val[0] == 1 && set[0])
			return 0L;
		dp[0][1] = 1;
		for(int i = 0; i < len; i++)
			for(int j = 0; j < len; j++){
				if(dp[i][j] == 0) continue;
				if(dp[i][j] < 0)
					dp[i][j] = INF;
				if(j == 0 && set[i + 1] && val[i + 1] == 1) continue;
				if(set[i + 1]){
					if(val[i + 1] == 0)
						dp[i + 1][j + 1] += dp[i][j];
					else
						dp[i + 1][j - 1] += dp[i][j];
				} else {
					dp[i + 1][j + 1] += dp[i][j];
					if(j > 0)
						dp[i + 1][j - 1] += dp[i][j];
				}
			}
		return dp[len - 1][0] >= 0 ? dp[len -1][0] : INF;
		
	}
	
	private void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);

		n = nextInt();
		m = nextInt();
		len = n+m - 1;
		long k = nextLong() - 1L;
		
		fill(pr, inf);
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++){
				pr[i + j] = min(pr[i + j], nextInt());
			}
		int pp[] = new int[len];
		for(int i = 0; i < len; i++)
			pp[i] = pr[i];
		sort(pp);
		
		for(int pos = 0; pos < len; pos++){
			int id = -1;
			for(int i = 0; i < len; i++)
				if(pr[i] == pp[pos])
					id = i;
			set[id] = true;
			val[id] = 0;
			long g = get();
			if(g > k) continue;
			else{
				val[id] = 1;
				k -= g;
			}
		}
		for(int i = 0; i < n; i++){
			for(int j = 0; j < m; j++){
				out.print(val[i + j] == 1 ? ')' : '(');
			}
			out.println();
		}
		
				
		
		in.close();
		out.close();
	}
	String nextToken() throws IOException {
		while (!st.hasMoreTokens())
			st = new StringTokenizer(in.readLine());
		return st.nextToken();
	}
	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}
	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	String nextLine() throws IOException {
		st = new StringTokenizer("");
		return in.readLine();
	}
	boolean EOF() throws IOException {
		while (!st.hasMoreTokens()) {
			String s = in.readLine();
			if (s == null)
				return true;
			st = new StringTokenizer(s);
		}
		return false;
	}
}
