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;
using ll = long long;
const ll INF = 1e18;
const int inf = 0x3f3f3f3f;
const int maxn = 105;
int n, m;
ll k;
int a[maxn][maxn];
int bracket[2 * maxn];
int mn[maxn * 2];
ll dp[205][105];
using pii = pair<int, int>;
priority_queue<pii, vector<pii>, greater<pii> > q;
ll solve() {
  dp[1][1] = 1;
  for (int i = 2; i <= n + m - 1; i++) {
    for (int j = 0; j <= 100; j++) {
      dp[i][j] = 0;
      if (bracket[i] == 0) {
        if (j >= 1) dp[i][j] = dp[i - 1][j - 1];
      } else if (bracket[i] == 1) {
        dp[i][j] = dp[i - 1][j + 1];
      } else {
        if (j >= 1) dp[i][j] += dp[i - 1][j - 1];
        dp[i][j] += dp[i - 1][j + 1];
        dp[i][j] = min(dp[i][j], INF);
      }
    }
  }
  return dp[n + m - 1][0];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  for (int i = 1; i <= m; i++) {
    mn[i] = inf;
    for (int j = 1; i - j + 1 >= 1 && j <= n; j++)
      mn[i] = min(mn[i], a[j][i - j + 1]);
  }
  for (int i = 2; i <= n; i++) {
    mn[m + i - 1] = inf;
    for (int j = 1; i + j - 1 <= n && m - j + 1 >= 1; j++)
      mn[m + i - 1] = min(mn[m + i - 1], a[i + j - 1][m - j + 1]);
  }
  for (int i = 2; i <= n + m - 2; i++) q.push({mn[i], i});
  memset(bracket, -1, sizeof bracket);
  bracket[1] = 0;
  bracket[n + m - 1] = 1;
  while (!q.empty()) {
    pii p = q.top();
    q.pop();
    bracket[p.second] = 0;
    ll num = solve();
    if (num >= k) continue;
    k -= num;
    bracket[p.second] = 1;
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; i - j + 1 >= 1 && j <= n; j++) a[j][i - j + 1] = bracket[i];
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 1; i + j - 1 <= n && m - j + 1 >= 1; j++)
      a[i + j - 1][m - j + 1] = bracket[m + i - 1];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (a[i][j] == 0)
        cout << "(";
      else
        cout << ")";
    }
    cout << "\n";
  }
  return 0;
}
