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;
template <class T>
T gcd(T a, T b) {
  return a == 0 ? b : gcd(b % a, a);
}
template <class T>
string tostring(T a) {
  ostringstream os;
  os << a;
  return os.str();
}
int toint(string a) {
  istringstream is(a);
  int p;
  is >> p;
  return p;
}
long long toll(string a) {
  istringstream is(a);
  long long p;
  is >> p;
  return p;
}
long long count(vector<int> v) {
  vector<vector<long long> > vv(v.size() + 1,
                                vector<long long>(v.size() + 1, 0));
  vv[0][0] = 1;
  for (int i = 1; i <= v.size(); i++) {
    for (int j = 0; j <= i - j - 1; j++) {
      if (v[i - 1] != 1) vv[j][i - j] += vv[j][i - j - 1];
      if (v[i - 1] != -1) vv[j + 1][i - j - 1] += vv[j][i - j - 1];
    }
  }
  return vv[v.size() / 2][v.size() / 2];
}
double count1(vector<int> v) {
  vector<vector<double> > vv(v.size() + 1, vector<double>(v.size() + 1, 0));
  vv[0][0] = 1;
  for (int i = 1; i <= v.size(); i++) {
    for (int j = 0; j <= i - j - 1; j++) {
      if (v[i - 1] != 1) vv[j][i - j] += vv[j][i - j - 1];
      if (v[i - 1] != -1) vv[j + 1][i - j - 1] += vv[j][i - j - 1];
    }
  }
  return vv[v.size() / 2][v.size() / 2] + 0.5;
}
int main() {
  long long m, n, k;
  cin >> m >> n >> k;
  int t;
  vector<int> v(m + n - 1, 20000);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      cin >> t;
      v[i + j] = min(v[i + j], t);
    }
  }
  vector<pair<int, int> > vp;
  for (int i = 0; i < v.size(); i++) {
    vp.push_back(make_pair(v[i], i));
  }
  sort(vp.begin(), vp.end());
  vector<int> vvvv(vp.size(), 0);
  for (int i = 0; i < vp.size(); i++) {
    vvvv[vp[i].second] = -1;
    long long ct = count(vvvv);
    double ct1 = count1(vvvv);
    if (ct1 < 2 * k && ct < k) {
      k -= (long long)ct;
      vvvv[vp[i].second] = 1;
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if (vvvv[i + j] == -1)
        cout << "(";
      else
        cout << ")";
    }
    cout << endl;
  }
}
