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 T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  os << "{" << p.first << ", " << p.second << "}";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[";
  int i = 0;
  for (typename vector<T>::const_iterator it = p.begin(); it != p.end(); ++it) {
    os << (i == 0 ? "" : ", ") << *it;
    i++;
  }
  os << "]";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "[";
  int i = 0;
  for (typename set<T>::const_iterator it = p.begin(); it != p.end(); ++it) {
    os << (i == 0 ? "" : ", ") << *it;
    i++;
  }
  os << "]";
  return os;
}
template <class K, class V>
ostream& operator<<(ostream& os, const map<K, V>& p) {
  os << "[";
  int i = 0;
  for (typename map<K, V>::const_iterator it = p.begin(); it != p.end(); ++it) {
    os << (i == 0 ? "" : ", ") << *it;
    i++;
  }
  os << "]";
  return os;
}
template <class T>
string toString(const T& o) {
  stringstream ss;
  ss << o;
  string ret;
  getline(ss, ret);
  return ret;
}
template <class T>
vector<T> subv(const vector<T>& v, int from, int to) {
  return vector<T>(v.begin() + min(max(0, from), (int)v.size()),
                   v.begin() + min(max(0, to), (int)v.size()));
}
pair<int, int> operator+(const pair<int, int>& a, const pair<int, int>& b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<int, int> operator*(const int& a, const pair<int, int>& b) {
  return make_pair(a * b.first, a * b.second);
}
long long MX = numeric_limits<long long>::max();
int n, m, N;
long long k;
vector<int> prior;
vector<char> ans;
long long dp[200][200];
void printAns() {
  for (int i = 0; i < N; ++i)
    if (ans[i] == 0)
      cout << '.';
    else
      cout << ans[i];
  cout << endl;
}
long long count() {
  memset(dp, 0, sizeof(dp));
  if (ans[0] == 0 || ans[0] == '(') dp[0][1] = 1;
  for (int pos = 1; pos < N; ++pos) {
    for (int bal = 0; bal <= N / 2; ++bal) {
      long long& v = dp[pos][bal];
      if (ans[pos] == 0 || ans[pos] == '(')
        if (bal - 1 >= 0) {
          long long val = dp[pos - 1][bal - 1];
          if (v == MX || val == MX)
            v = MX;
          else
            v += val;
        }
      if (ans[pos] == 0 || ans[pos] == ')') {
        long long val = dp[pos - 1][bal + 1];
        if (v == MX || val == MX)
          v = MX;
        else
          v += val;
      }
      if (dp[pos][bal] < 0) dp[pos][bal] = MX;
    }
  }
  long long ret = dp[N - 1][0];
  return ret;
}
void solve() {
  for (int i = 0; i < ans.size(); ++i) {
    int ind = 0;
    for (int j = 0; j < prior.size(); ++j)
      if (prior[ind] > prior[j]) ind = j;
    prior[ind] = INT_MAX;
    ans[ind] = '(';
    long long cnt = count();
    if (k > cnt) {
      k -= cnt;
      ans[ind] = ')';
    }
  }
}
int main() {
  cin >> n >> m >> k;
  N = n + m - 1;
  prior.resize(N, INT_MAX);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      int p;
      cin >> p;
      prior[i + j] = min(prior[i + j], p);
    }
  ans.resize(N, 0);
  solve();
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) cout << ans[i + j];
    cout << endl;
  }
  return 0;
}
