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 <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
string ans = "";
long long dp[201][201];
long long fun2(long long x, long long y) {
  if (x >= 1000000000000000000 || y >= 1000000000000000000)
    return 1000000000000000000;
  return x + y;
}
long long fun1(long long n, long long ind) {
  ans[ind] = '(';
  memset(dp, 0, sizeof(dp));
  dp[1][1] = 1;
  for (int i = 2; i < n + 1; i++) {
    for (int j = 0; j < n + 1; j++) {
      if ((ans[i - 1] == '(' || ans[i - 1] == '*') && j - 1 >= 0)
        dp[i][j] = fun2(dp[i][j], dp[i - 1][j - 1]);
      if ((ans[i - 1] == ')' || ans[i - 1] == '*') && j + 1 <= n)
        dp[i][j] = fun2(dp[i][j], dp[i - 1][j + 1]);
    }
  }
  ans[ind] = '*';
  return dp[n][0];
}
void fun(long long n, long long c[], long long k) {
  vector<pair<long long, long long> > temp;
  for (int i = 0; i < n; i++) temp.push_back(make_pair(c[i], i));
  sort(temp.begin(), temp.end());
  for (int i = 0; i < n; i++) ans = ans + '*';
  for (int i = 0; i < n; i++) {
    long long temp1 = fun1(n, temp[i].second);
    if (temp1 < k) {
      k -= temp1;
      ans[temp[i].second] = ')';
    } else
      ans[temp[i].second] = '(';
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  long long a[n][m];
  long long c[n + m - 1];
  memset(c, -1, sizeof(c));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      if (c[i + j] == -1 || c[i + j] > a[i][j]) c[i + j] = a[i][j];
    }
  fun(n + m - 1, c, k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cout << ans[i + j];
    cout << endl;
  }
  return 0;
}
