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.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main {
    private static int[] p;
    private static Integer[] rk;
    private static long[][] dp;
    private static char[] s;

    private static class Comp implements Comparator<Integer> {

        @Override
        public int compare(Integer lhs, Integer rhs) {
            return p[lhs] - p[rhs];
        }
    }

    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);

        int r = cin.nextInt();
        int c = cin.nextInt();
        long k = cin.nextLong();
        int n = r + c - 1;

        p = new int[n];
        rk = new Integer[n];
        dp = new long[n + 1][n + 1];
        s = new char[n + 1];
        Arrays.fill(s, '\0');
        for (int i = 0; i < n; i++) {
            rk[i] = i;
            p[i] = 0x3f3f3f3f;
        }
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                int x = cin.nextInt();
                p[i + j] = Math.min(p[i + j], x);
            }
        }

        Arrays.sort(rk, 0, n, new Comp());

        for (int id = 0; id < n; id++) {
            s[rk[id]] = '(';
            for (int i = 0; i <= n; i++)
                Arrays.fill(dp[i], 0);
            dp[0][0] = 1;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j <= i && j <= n - i; j++) {
                    if (dp[i][j] != 0) {
                        if (dp[i][j] > k)
                            dp[i][j] = k;
                        if (s[i] != ')')
                            dp[i + 1][j + 1] += dp[i][j];
                        if (s[i] != '(' && j != 0)
                            dp[i + 1][j - 1] += dp[i][j];
                    }
                }
            }
            if (k > dp[n][0]) {
                k -= dp[n][0];
                s[rk[id]] = ')';
            }
        }

        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                System.out.print(s[i + j]);
            }
            System.out.println();
        }
    }
}
