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>
const long long MAX_N = 110 + 100;
const long long MOD = 1e9 + 7;
using namespace std;
long long dp[MAX_N][MAX_N];
int t[MAX_N][MAX_N];
int a[MAX_N];
int b[MAX_N];
long long n, m, k, len;
void solve() {
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < MAX_N; ++i) dp[0][i] = 1;
  for (int i = 2; i <= len; i += 2) {
    for (int j = 1; j <= len - i + 1; ++j) {
      if (a[j] == 2 or a[j + i - 1] == 1) continue;
      for (int x = 2; x <= i; x += 2) {
        if (a[j + x - 1] == 1) continue;
        if (x < i) {
          long long f;
          long long u = dp[x - 2][j + 1];
          long long v = dp[i - x][j + x];
          if (u and v and (k / (long double)u) < v)
            f = k + 1;
          else
            f = u * v;
          dp[i][j] += f;
          if (dp[i][j] > k) dp[i][j] = k + 1;
        } else {
          dp[i][j] += dp[i - 2][j + 1];
          if (dp[i][j] > k) dp[i][j] = k + 1;
        }
      }
    }
  }
}
bool check() {
  int now = 0;
  int c = 0;
  int x[len + 1];
  for (int i = 1; i <= len; ++i) x[i] = a[i];
  for (int i = 1; i <= len; ++i) c += (a[i] == 1);
  c = (len / 2) - c;
  bool ok = true;
  for (int i = 1; i <= len and c; ++i) {
    if (!a[i]) a[i] = 1, c--;
  }
  for (int i = 1; i <= len; ++i) {
    now += (a[i] == 1);
    now -= (a[i] != 1);
    if (now < 0) ok = false;
  }
  if (now) ok = false;
  for (int i = 1; i <= len; ++i) a[i] = x[i];
  return ok;
}
int main() {
  cin >> n >> m >> k;
  k--;
  len = m + n - 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) cin >> t[i][j];
  for (int i = 2; i <= len + 1; ++i) {
    int mn = 1e9;
    for (int x = 1; x <= i - 1; x++) {
      if (x <= n and i - x <= m) mn = min(mn, t[x][i - x]);
    }
    b[i - 1] = mn;
  }
  a[1] = 1;
  a[len] = 2;
  set<pair<int, int>> st;
  for (int i = 2; i <= len - 1; ++i) st.insert({b[i], i});
  while (st.size() and k) {
    int i = (*st.begin()).second;
    a[i] = 1;
    st.erase(st.begin());
    if (!check()) {
      a[i] = 2;
      continue;
    }
    solve();
    if (k - dp[len][1] >= 0) {
      a[i] = 2;
      k -= dp[len][1];
    }
  }
  while (st.size()) {
    int i = (*st.begin()).second;
    a[i] = 1;
    st.erase(st.begin());
    if (!check()) a[i] = 2;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (a[i + j - 1] == 1)
        cout << "(";
      else
        cout << ")";
    }
    cout << "\n";
  }
  return 0;
}
