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 MAX_N = 103;
const int MAX_M = 103;
const int MAX_A = MAX_N + MAX_M;
const long long oo = 1000000000000000099LL;
int n, m, nA;
pair<pair<int, int>, pair<int, int> > a[MAX_A];
int b[MAX_A], p[MAX_N][MAX_N], res[MAX_N][MAX_N];
long long k;
long long f[MAX_A][MAX_A];
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
long long solve(int p, int v) {
  b[p] = v;
  memset(f, (0), sizeof(f));
  f[0][1] = 1;
  for (int i = int(1); i <= int(nA); ++i)
    if (b[i] == 0) {
      f[i][0] = f[i - 1][1];
      for (int j = int(1); j <= int(nA); ++j) {
        f[i][j] += f[i - 1][j - 1] + f[i - 1][j + 1];
        if (f[i][j] > k) f[i][j] = oo;
      }
    } else {
      if (b[i] == -1) f[i][0] = f[i - 1][1];
      for (int j = int(1); j <= int(nA); ++j) {
        f[i][j] += f[i - 1][j - b[i]];
        if (f[i][j] > k) f[i][j] = oo;
      }
    }
  b[p] = 0;
  return f[nA][1];
}
int get_min(int i, int j) {
  if (i < 1 || j < 1 || i > n || j > m) return 1000000000;
  return min(p[i][j], get_min(i + 1, j - 1));
}
void copy_res(int i, int j, int v) {
  if (i < 1 || j < 1 || i > n || j > m) return;
  res[i][j] = v;
  copy_res(i + 1, j - 1, v);
}
int main() {
  cin >> n >> m >> k;
  for (int i = int(1); i <= int(n); ++i)
    for (int j = int(1); j <= int(m); ++j) scanf("%d", &p[i][j]);
  nA = 0;
  for (int j = int(2); j <= int(m - int(n == 1)); ++j) {
    a[nA] = make_pair((make_pair((get_min(1, j)), (nA + 1))),
                      (make_pair((1), (j))));
    ++nA;
  }
  for (int i = int(2); i <= int(n - 1); ++i) {
    a[nA] = make_pair((make_pair((get_min(i, m)), (nA + 1))),
                      (make_pair((i), (m))));
    ++nA;
  }
  sort(a, a + nA);
  if (n == 44) {
  }
  memset(b, (0), sizeof(b));
  memset(res, (0), sizeof(res));
  for (int i = 0; i < int(nA); ++i) {
    int j = a[i].first.second;
    long long t = solve(j, 1);
    if (t > 0 && t >= k) {
      b[j] = res[a[i].second.first][a[i].second.second] = 1;
    } else {
      b[j] = res[a[i].second.first][a[i].second.second] = -1;
      k -= t;
    }
  }
  res[1][1] = 1;
  res[n][m] = -1;
  for (int j = int(2); j <= int(m); ++j) copy_res(2, j - 1, res[1][j]);
  for (int i = int(2); i <= int(n - 1); ++i) copy_res(i + 1, m - 1, res[i][m]);
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(1); j <= int(m); ++j)
      if (res[i][j] == 1)
        printf("(");
      else
        printf(")");
    printf("\n");
  }
  return 0;
}
