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:64000000")
using namespace std;
void __never(int a) { printf("\nOPS %d", a); }
void __die(int a) { printf("%d", (a * a) / (a + a)); }
long long k;
int n, m;
int p[128][128];
bool used[128 + 128];
long long INF = (long long)2000000000 * 1000000000;
long long dp[128 + 128][128 + 128];
long long sol3(string s) {
  dp[0][0] = 1;
  int n = (int)((s).size());
  for (int(i) = (1); (i) <= (n); ++(i)) {
    char ch = s[i - 1];
    for (int(j) = (0); (j) <= (i); ++(j)) {
      if (ch == '(')
        dp[i][j] = j > 0 ? dp[i - 1][j - 1] : 0;
      else if (ch == ')')
        dp[i][j] = dp[i - 1][j + 1];
      else
        dp[i][j] = (j > 0 ? dp[i - 1][j - 1] : 0) + dp[i - 1][j + 1];
      if (dp[i][j] > INF) dp[i][j] = INF;
    }
  }
  return dp[n][0];
}
string sol2(vector<int> p, long long k) {
  int n = (int)((p).size());
  string ans(n, '?');
  memset(used, 0, sizeof(used));
  for (int(I) = (1); (I) <= (n); ++(I)) {
    int mi = -1;
    int ind = -1;
    for (int(i) = (0); (i) <= (n - 1); ++(i))
      if (!used[i] && (p[i] < mi || mi == -1)) {
        ind = i;
        mi = p[i];
      }
    used[ind] = true;
    ans[ind] = '(';
    long long tt = sol3(ans);
    if (tt < k) {
      k -= tt;
      ans[ind] = ')';
    }
  }
  return ans;
}
void init() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void sol() {
  cin >> n >> m >> k;
  for (int(i) = (1); (i) <= (n); ++(i))
    for (int(j) = (1); (j) <= (m); ++(j)) scanf("%d", &p[i][j]);
  for (int(i) = (n); (i) >= (1); --(i))
    for (int(j) = (1); (j) <= (m); ++(j)) {
      if (i < n && j > 1) p[i][j] = min(p[i][j], p[i + 1][j - 1]);
    }
  vector<int> p2(n + m - 1);
  for (int(i) = (1); (i) <= (m); ++(i)) p2[i - 1] = p[1][i];
  for (int(i) = (2); (i) <= (n); ++(i)) p2[i - 2 + m] = p[i][m];
  string ans2 = sol2(p2, k);
  for (int(i) = (0); (i) <= (n - 1); ++(i)) cout << ans2.substr(i, m) << endl;
}
int main() {
  sol();
  return 0;
}
