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:64000000")
using namespace std;
const int INF = ((1 << 31) - 1);
const long long LLINF = (((1LL << 63) - 1LL));
const double eps = 1e-9;
const double PI = 3.14159265358979323846;
long long add(long long a, long long b) {
  long long B = 1e18 + 100;
  return min(min(a, B) + min(b, B), B);
}
int main() {
  int n, m;
  long long k;
  cin >> n >> m;
  cin >> k;
  assert((n + m) & 1);
  vector<vector<int> > p(n, vector<int>(m, -1));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      scanf("%d", &p[i][j]);
    }
  }
  int len = n + m - 1;
  vector<int> prior(len, INF);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      prior[i + j] = min(prior[i + j], p[i][j]);
    }
  vector<int> perm;
  for (int i = 0; i < len; ++i) {
    int id = 0;
    for (int j = 0; j < len; ++j)
      if (prior[j] < prior[id]) id = j;
    perm.push_back(id);
    prior[id] = INF;
  }
  string s(len, '?');
  --k;
  for (int i = 0; i < len; ++i) {
    int pos = perm[i];
    s[pos] = '(';
    vector<vector<long long> > dp(len + 1, vector<long long>(len + 1, 0));
    dp[0][len] = 1;
    for (int i = len - 1; i >= 0; --i) {
      for (int h = 0; h <= len; ++h) {
        long long open = 0;
        long long close = 0;
        if (h != len) open = add(open, dp[h + 1][i + 1]);
        if (h != 0) close = add(close, dp[h - 1][i + 1]);
        if (s[i] == '?') {
          dp[h][i] = add(open, close);
        } else if (s[i] == '(') {
          dp[h][i] = open;
        } else if (s[i] == ')') {
          dp[h][i] = close;
        }
      }
    }
    long long amount = dp[0][0];
    if (amount - 1 < k) {
      k -= amount;
      s[pos] = ')';
    }
  }
  cerr << k;
  if (k != 0) {
    long long* fail = new long long[1 << 27];
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      printf("%c", s[i + j]);
    }
    printf("\n");
  }
  return 0;
}
