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>
#pragma comment(linker, "/STACK:300000000")
using namespace std;
const double pi = 3.1415926535897932384626433832795;
template <class T>
T myabs(T a) {
  return a >= 0 ? a : -a;
};
template <class T>
T sqr(T x) {
  return x * x;
}
int n, m;
long long k, curK;
string ans;
vector<vector<int> > p;
long long dp[210][210];
double dpAppr[210][210];
double d;
long long inf = 2LL * 1000000000 * 1000000000;
long long getCnt(string &s) {
  memset(dp, 0, sizeof(dp));
  memset(dpAppr, 0, sizeof(dpAppr));
  dp[0][0] = 1;
  dpAppr[0][0] = 1;
  for (int i = 1; i <= ((int)(s).size()); i++) {
    for (int j = 0; j < ((int)(s).size()) / 2 + 1; j++) {
      if (s[i - 1] != '(') {
        dp[i][j] += dp[i - 1][j + 1];
        dpAppr[i][j] += dpAppr[i - 1][j + 1];
      }
      if (s[i - 1] != ')') {
        if (j != 0) {
          dp[i][j] += dp[i - 1][j - 1];
          dpAppr[i][j] += dpAppr[i - 1][j - 1];
        }
      }
    }
  }
  if (dpAppr[((int)(s).size())][0] >= inf) return inf;
  return dp[((int)(s).size())][0];
}
int cx, cy;
void getCell() {
  int mi = 999999;
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) {
      if (p[i][j] < mi) {
        mi = p[i][j];
        cx = i;
        cy = j;
      }
    }
}
int main() {
  cin >> n >> m >> k;
  k--;
  curK = 0;
  ans.resize(n + m - 1, '?');
  p.resize(n, vector<int>(m));
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) cin >> p[i][j];
  for (int i = 0; i < (n * m); i++) {
    getCell();
    p[cx][cy] = 999999;
    int ind = cx + cy;
    if (ans[ind] != '?') continue;
    ans[ind] = '(';
    long long t = getCnt(ans);
    if (curK + t <= k) {
      curK += t;
      ans[ind] = ')';
    }
  }
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (m); j++) cout << ans[i + j];
    cout << endl;
  }
}
