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;
long long vx[4] = {1, 0, -1, 0};
long long vy[4] = {0, 1, 0, -1};
long long dp[210][210];
long long vis[210][210];
long long calc(char* str, long long pos, long long depth, long long k) {
  if (vis[pos][depth]) return dp[pos][depth];
  vis[pos][depth] = 1;
  long long res = 0, tmp;
  if (depth >= 0) {
    if (pos > 0) {
      if (depth <= pos) {
        if (str[pos - 1] != '(') {
          tmp = calc(str, pos - 1, depth + 1, k);
          if (tmp < 0) return dp[pos][depth] = -1;
          res += tmp;
        }
        if (str[pos - 1] != ')') {
          tmp = calc(str, pos - 1, depth - 1, k);
          if (tmp < 0) return dp[pos][depth] = -1;
          res += tmp;
        }
      }
    } else {
      if (depth == 0) {
        res = 1;
      }
    }
  }
  if (res >= k) res = -1;
  ;
  return dp[pos][depth] = res;
}
int main() {
  long long n, m, k, p[105][105];
  cin >> n >> m >> k;
  for (long long i = 0; i < ((long long)n); i++)
    for (long long j = 0; j < ((long long)m); j++) cin >> p[i][j];
  long long minp[210];
  for (long long i = 0; i < ((long long)210); i++) minp[i] = 3 * m * n;
  for (long long d = 0; d <= n + m - 2; d++) {
    for (long long i = 0; i < n; i++) {
      long long j = d - i;
      if (0 <= j && j < m) {
        minp[d] = min(minp[d], p[i][j]);
      }
    }
  }
  vector<pair<long long, long long> > prio;
  for (long long i = 0; i < ((long long)n + m - 1); i++)
    prio.push_back(make_pair(minp[i], i));
  sort(prio.begin(), prio.end());
  char ans[210];
  for (long long i = 0; i < ((long long)210); i++) ans[i] = '*';
  ans[0] = '(';
  ans[n + m - 2] = ')';
  for (long long j = 0; j < ((long long)prio.size()); j++) {
    long long i = prio[j].second;
    if (i == 0 || i == n + m - 2) continue;
    ans[i] = '(';
    for (long long a = 0; a < ((long long)210); a++)
      for (long long b = 0; b < ((long long)210); b++) vis[a][b] = 0;
    long long res = calc(ans, n + m - 1, 0, k);
    if (0 <= res && res < k) {
      ans[i] = ')';
      k -= res;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cout << ans[i + j];
    }
    cout << endl;
  }
  return 0;
}
