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 int N = 100 + 100;
const long long int mod = 1e9 + 7;
const long long int inf = 1e18 + 100;
long long int a[N][N], ans[N], dp[N][N];
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 0; i < m; i++) {
    long long int mn = a[0][i];
    int x = 0, y = i;
    while (x + 1 < n && y - 1 >= 0) {
      x++;
      y--;
      mn = min(mn, a[x][y]);
    }
    v.push_back(mn);
  }
  for (int i = 1; i < n; i++) {
    int x = i, y = m - 1;
    long long int mn = a[x][y];
    while (x + 1 < n && y - 1 >= 0) {
      x++;
      y--;
      mn = min(mn, a[x][y]);
    }
    v.push_back(mn);
  }
  int sz = v.size();
  int sz2 = sz;
  for (int i = 0; i < sz; i++) ans[i] = -1;
  while (sz2--) {
    int mn = -1;
    for (int j = 0; j < sz; j++) {
      if (ans[j] == -1 && (mn == -1 || v[j] < v[mn])) mn = j;
    }
    ans[mn] = 1;
    for (int i = 0; i < sz; i++) {
      for (int j = 0; j <= sz; j++) {
        dp[i][j] = 0;
      }
    }
    if (ans[0] == -1 || ans[0] == 1) dp[0][1] = 1;
    for (int i = 1; i < sz; i++) {
      for (int j = 0; j <= sz; j++) {
        if (ans[i] == -1 || ans[i] == 0) dp[i][j] = dp[i - 1][j + 1];
        if (j != 0 && (ans[i] == -1 || ans[i] == 1))
          dp[i][j] += dp[i - 1][j - 1];
        dp[i][j] = min(inf, dp[i][j]);
      }
    }
    if (k > dp[sz - 1][0]) {
      k -= dp[sz - 1][0];
      ans[mn] = 0;
    }
  }
  int ind = 0;
  for (int i = 0; i < m; i++) {
    int x = 0, y = i;
    a[x][y] = ans[ind];
    while (x + 1 < n && y - 1 >= 0) {
      x++;
      y--;
      a[x][y] = ans[ind];
    }
    ind++;
  }
  for (int i = 1; i < n; i++) {
    int x = i, y = m - 1;
    a[x][y] = ans[ind];
    while (x + 1 < n && y - 1 >= 0) {
      x++;
      y--;
      a[x][y] = ans[ind];
    }
    ind++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j] == 1)
        cout << '(';
      else
        cout << ')';
    }
    cout << endl;
  }
  return 0;
}
