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>
using namespace std;
const long double epsilon = 1e-9;
long long cnt[256][128];
long long infi = 1LL << 60;
long long count(const string& res) {
  memset(cnt, 0, sizeof(cnt));
  cnt[0][0] = 1;
  int n = res.size() / 2;
  for (int i = 0; i < res.size(); i++) {
    if (res[i] == '?') {
      for (int j = 0; j <= n; j++) {
        if (j > 0) {
          cnt[i + 1][j - 1] = min(infi, cnt[i + 1][j - 1] + cnt[i][j]);
        }
        if (j < n) cnt[i + 1][j + 1] = min(infi, cnt[i + 1][j + 1] + cnt[i][j]);
      }
    } else if (res[i] == '(') {
      for (int j = 0; j < n; j++) {
        cnt[i + 1][j + 1] = min(infi, cnt[i + 1][j + 1] + cnt[i][j]);
      }
    } else {
      for (int j = 1; j <= n; j++) {
        cnt[i + 1][j - 1] = min(infi, cnt[i + 1][j - 1] + cnt[i][j]);
      }
    }
  }
  return cnt[2 * n][0];
}
int mat[128][128];
int main() {
  int n, m;
  long long k;
  cin >> n >> m >> k;
  vector<pair<int, int> > v(n * m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &mat[i][j]);
      v[mat[i][j] - 1] = make_pair(i, j);
    }
  }
  vector<bool> placed(n + m - 1, false);
  vector<int> priorities;
  for (int i = 0; i < v.size(); i++) {
    if (placed[v[i].first + v[i].second]) continue;
    placed[v[i].first + v[i].second] = true;
    priorities.push_back(v[i].first + v[i].second);
  }
  string helper;
  helper.resize(n + m - 1, '?');
  long long t;
  for (int i = 0; i < priorities.size(); i++) {
    helper[priorities[i]] = '(';
    if (k > (t = count(helper))) {
      k -= t;
      helper[priorities[i]] = ')';
    }
  }
  vector<string> ans(n);
  for (int i = 0; i < n; i++) {
    ans[i].resize(m);
    for (int j = 0; j < m; j++) ans[i][j] = helper[i + j];
    cout << ans[i] << endl;
  }
  return 0;
}
