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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {

    static BufferedReader bf = new BufferedReader(new InputStreamReader(
            System.in));
    static StringTokenizer st;
    static PrintWriter out = new PrintWriter(System.out);

    static String nextToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            String s = bf.readLine();
            if (s == null)
                return null;
            st = new StringTokenizer(s);
        }

        return st.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    static String nextStr() throws IOException {
        return nextToken();
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
    
    static long f(char s[], long k, long a[][], int d, int p) {
        if (d > p + 1) {
            return 0;
        }
        if (p == 0) {
            if (d == 1) {
                return 1;
            } else {
                return 0;
            }
        }
        
        if (a[d][p] != Long.MAX_VALUE) {
            return a[d][p];
        }
        
        long res = 0;
        if (s[p] == '(') {
            if (d != 0) {
                res = f(s, k, a, d - 1, p - 1);
            }
        } else if (s[p] == ')') {
            res = f(s, k, a, d + 1, p - 1);
        } else {
            if (d != 0) {
                res = f(s, k, a, d - 1, p - 1) + f(s, k, a, d + 1, p - 1);
            } else {
                res = f(s, k, a, d + 1, p - 1);
            }
        }
        
        if (res > k) {
            a[d][p] = k + 1;
        } else {
            a[d][p] = res;
        }
        
        return a[d][p];
    }

    public static void main(String[] args) throws IOException {
        int n = nextInt();
        int m = nextInt();
        long k = nextLong();
        
        long p[][] = new long[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                p[i][j] = nextLong();
            }
        }
        
        int ord[] = new int[n + m - 1];
        int s = 1;
        for (int t = 0; t < m * n; t++) {
            int x = -1;
            int y = -1;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (p[i][j] != Long.MAX_VALUE && (x == -1 || p[i][j] < p[x][y])) {
                        x = i;
                        y = j;
                    }
                }
            }
            
            if (ord[x + y] == 0) {
                ord[x + y] = s;
                s++;
            }
            p[x][y] = Long.MAX_VALUE;
        }
        
        int l = n + m - 1;
        char res[] = new char[n + m - 1];
        res[0] = '(';
        res[n + m - 2] = ')';
        
        long a[][] = new long[l][l];
        
        for (int q = 0; q < l - 2; q++) {
            for (int i = 0; i < l; i++) {
                for (int j = 0; j < l; j++) {
                    a[i][j] = Long.MAX_VALUE;
                }
            }
            
            int t = -1;
            for (int i = 0; i < l; i++) {
                if (res[i] == 0 && (t == -1 || ord[t] > ord[i])) {
                    t = i;
                }
            }
            
            res[t] = '(';
            long tmp = f(res, k, a, 0, l - 1);
            if (tmp < k) {
                k -= tmp;
                res[t] = ')';
            }
        }
        
        char r[][] = new char[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                r[i][j] = res[i + j];
            }
            out.println(new String(r[i]));
        }
        out.flush();
        
    }
}
