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;
void require(bool cond, const string& message = "Runtime error") {
  if (!cond) {
    cerr << message << endl;
    assert(false);
  }
}
long long k;
vector<int> ord;
int N, M;
map<pair<int, int>, int> pos;
void readData() {
  int n, m;
  cin >> n >> m >> k;
  N = n;
  M = m;
  vector<vector<int> > a(n, vector<int>(m));
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) cin >> a[i][j];
  vector<int> p(n + m - 1, n * m + 1);
  for (int i = 0; i < int(p.size()); ++i) {
    int x = i, y = 0;
    while (x >= 0) {
      if (x < n && y < m) {
        p[i] = min(p[i], a[x][y]);
        pos[make_pair(x, y)] = i;
      }
      --x;
      ++y;
    }
  }
  vector<pair<int, int> > vp;
  for (int i = 0; i < int(p.size()); ++i) vp.push_back(make_pair(p[i], i));
  sort(vp.begin(), vp.end());
  for (int i = 0; i < int(vp.size()); ++i) ord.push_back(vp[i].second);
}
long long calc(string s, long long k) {
  int n = s.size();
  vector<vector<long long> > d(n + 1, vector<long long>(n + 1, 0));
  const long long INF = 1000000000ll * 1000000000ll + 10;
  d[0][0] = 1;
  for (int i = 0; i < int(n); ++i) {
    for (int j = 0; j < int(i + 1); ++j) {
      if (s[i] != ')') {
        d[i + 1][j + 1] += d[i][j];
        if (d[i + 1][j + 1] > INF) d[i + 1][j + 1] = INF;
      }
      if (s[i] != '(' && j > 0) {
        d[i + 1][j - 1] += d[i][j];
        if (d[i + 1][j - 1] > INF) d[i + 1][j - 1] = INF;
      }
    }
  }
  return d[n][0];
}
void solve() {
  int n = ord.size();
  string s(n, '.');
  --k;
  for (int i = 0; i < int(ord.size()); ++i) {
    s[ord[i]] = '(';
    long long cnt = calc(s, k);
    if (k >= cnt) {
      k -= cnt;
      s[ord[i]] = ')';
    }
  }
  vector<string> ans(N, string(M, '.'));
  for (int x = 0; x < int(N); ++x) {
    for (int y = 0; y < int(M); ++y) {
      ans[x][y] = s[pos[make_pair(x, y)]];
    }
  }
  for (int x = 0; x < int(N); ++x) {
    cout << ans[x] << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  readData();
  solve();
  return 0;
}
