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:200000000")
using namespace std;
const int INF = (int)1E9;
const long long INF64 = (long long)2E18;
const double EPS = 1E-9;
const double PI = acos((double)0) * 2;
const int MAXN = 210;
int n, m, p[MAXN][MAXN];
long long k;
bool read() {
  if (!(cin >> n >> m >> k)) return false;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) scanf("%d", &p[i][j]);
  return true;
}
int prior[MAXN];
pair<int, int> vec[MAXN];
string pat;
int cu, u[MAXN][MAXN];
long long d[MAXN][MAXN];
long long get_d(int pos, int bal) {
  if (pos == (int)pat.length()) return bal == 0;
  int& myu = u[pos][bal];
  long long& my = d[pos][bal];
  if (myu == cu) return my;
  my = 0;
  myu = cu;
  if (pat[pos] != ')') my += get_d(pos + 1, bal + 1);
  if (bal && pat[pos] != '(') my += get_d(pos + 1, bal - 1);
  my = min(my, INF64);
  return my;
}
long long calc() {
  ++cu;
  return get_d(0, 0);
}
void rec(int pos1, long long k) {
  if (pos1 == n + m - 1) {
    for (int i = 0; i < int(n); ++i) {
      for (int j = 0; j < int(m); ++j) printf("%c", pat[i + j]);
      puts("");
    }
    return;
  }
  int pos = vec[pos1].second;
  pat[pos] = '(';
  long long cur = calc();
  if (cur >= k) return rec(pos1 + 1, k);
  k -= cur;
  pat[pos] = ')';
  if (calc() < k) throw;
  return rec(pos1 + 1, k);
}
void solve() {
  for (int i = 0; i < int(n + m - 1); ++i) prior[i] = INF;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) prior[i + j] = min(prior[i + j], p[i][j]);
  for (int i = 0; i < int(n + m - 1); ++i) vec[i] = make_pair(prior[i], i);
  sort(vec, vec + n + m - 1);
  pat = string(n + m - 1, '?');
  rec(0, k);
}
int main() {
  if (!read()) throw;
  solve();
}
