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;
using namespace rel_ops;
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const long long inf = 2000000000000000000LL;
struct CF123C {
  long long count(string& preset) {
    int n = ((int)(preset).size());
    vector<vector<long long> > cnt_lvl(n, vector<long long>(n + 10));
    if (preset[0] == ')') return 0;
    cnt_lvl[0][1] = 1;
    for (int i = 1; i < n; i++) {
      for (int j = 0; j < n + 5; j++) {
        if (j && preset[i] != ')') cnt_lvl[i][j] += cnt_lvl[i - 1][j - 1];
        if (preset[i] != '(') cnt_lvl[i][j] += cnt_lvl[i - 1][j + 1];
        cnt_lvl[i][j] = min(cnt_lvl[i][j], inf);
      }
    }
    return cnt_lvl[n - 1][0];
  }
  string solve1(vector<int>& prio, long long k) {
    int n = ((int)(prio).size());
    vector<pair<int, int> > prio1(n);
    for (int i = 0; i < (n); i++) prio1[i] = make_pair(prio[i], i);
    sort((prio1).begin(), (prio1).end());
    string t;
    for (int i = 0; i < (n); i++) t += '.';
    for (typeof((prio1).begin()) it = (prio1).begin(); it != (prio1).end();
         ++it) {
      assert(k < count(t));
      int pos = (*it).second;
      t[pos] = '(';
      long long cnt = count(t);
      if (k >= cnt) {
        k -= cnt;
        t[pos] = ')';
      }
    }
    return t;
  }
  void solve(int n, int m, long long k, vector<vector<int> >& prio) {
    vector<int> prio1(n + m - 1, 1010101010);
    for (int i = 0; i < (n); i++)
      for (int j = 0; j < (m); j++)
        prio1[i + j] = min(prio1[i + j], prio[i][j]);
    string s = solve1(prio1, k);
    for (int i = 0; i < (n); i++) {
      for (int j = 0; j < (m); j++) cout << s[i + j];
      puts("");
    }
  }
};
int main() {
  int n, m;
  long long k;
  cin >> n >> m >> k;
  k--;
  vector<vector<int> > prio(n, vector<int>(m));
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) cin >> prio[i][j];
  CF123C x;
  x.solve(n, m, k, prio);
}
