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.
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
const unsigned long long int INF = 2e18;
unsigned long long int p[1000][1000];
int main() {
  unsigned long long int n, m, need;
  cin >> n >> m >> need;
  for (unsigned long long int i = 0; i < n; i++) {
    for (unsigned long long int j = 0; j < m; j++) {
      cin >> p[i][j];
    }
  }
  unsigned long long int len = n + m - 1;
  vector<pair<unsigned long long int, unsigned long long int> > diag(len);
  for (unsigned long long int k = 0; k <= n + m - 2; k++) {
    diag[k].first = INF;
    diag[k].second = k;
    for (unsigned long long int i = 0; i < n; i++) {
      if (k - i >= m || k - i < 0) continue;
      diag[k].first = min(diag[k].first, p[i][k - i]);
    }
  }
  sort(diag.begin(), diag.end());
  vector<unsigned long long int> ch(len, 2);
  unsigned long long int bal = 0;
  for (unsigned long long int i = 0; i < len; i++) {
    unsigned long long int now = diag[i].second;
    vector<unsigned long long int> f(len + 10, 0), s(len + 1, 0);
    f[0] = 1;
    bool can = false;
    unsigned long long int add = bal + 1;
    for (unsigned long long int j = 0; j < len; j++) {
      vector<unsigned long long int> newo = f;
      for (unsigned long long int k = 0; k <= len; k++) {
        if (ch[j] == 0 || j == now)
          newo[k] = (k ? f[k - 1] : 0), add--;
        else if (ch[j] == 2)
          newo[k] = (k ? f[k - 1] : 0) + f[k + 1];
        else if (ch[j] == 1)
          newo[k] = f[k + 1], add++;
        newo[k] = min(newo[k], INF);
        if (len - 1 - j == k + add && newo[k] >= need) {
        }
      }
      if (can) break;
      f = newo;
    }
    unsigned long long int det = f[0];
    if (can || det >= need) {
      ch[now] = 0;
      bal++;
    } else {
      ch[now] = 1;
      bal--;
      need -= det;
    }
  }
  for (unsigned long long int i = 0; i < n; i++) {
    for (unsigned long long int j = 0; j < m; j++) {
      cout << (ch[i + j] ? ')' : '(');
    }
    cout << endl;
  }
}
