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;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
long long mypow(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 1ll;
  long long _tmp = mypow(_a, _x / 2, _mod);
  _tmp = (_tmp * _tmp) % _mod;
  if (_x & 1) _tmp = (_tmp * _a) % _mod;
  return _tmp;
}
inline long long add(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x + _y;
  if (_ >= _mod) _ -= _mod;
  return _;
}
inline long long sub(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x - _y;
  if (_ < 0) _ += _mod;
  return _;
}
inline long long mul(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x * _y;
  if (_ >= _mod) _ %= _mod;
  return _;
}
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
int __ = 1, _cs;
long long n, m, k, nm, pp[111 + 111];
char aa[111 + 111];
void build() {}
vector<pair<int, int> > v;
void init() {
  n = getint();
  m = getint();
  k = getint();
  nm = n + m - 1;
  for (long long i = 1; i <= nm; i++) pp[i] = n * m;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++)
      pp[i + j - 1] = min(pp[i + j - 1], getint());
  for (long long i = 1; i <= nm; i++) v.push_back(make_pair(pp[i], i));
  sort((v).begin(), (v).end());
}
long long dp[111 + 111][111 + 111];
long long DP() {
  dp[0][0] = 1;
  for (long long i = 1; i <= nm; i++)
    for (long long j = 0; j <= i; j++) {
      dp[i][j] = 0;
      if (aa[i] != ')' && j) dp[i][j] += dp[i - 1][j - 1];
      if (aa[i] != '(' && j + 1 <= i - 1) dp[i][j] += dp[i - 1][j + 1];
      if (dp[i][j] > k) dp[i][j] = k + 1;
    }
  return dp[nm][0];
}
void solve() {
  for (size_t i = 0; i < v.size(); i++) {
    int p = v[i].second;
    aa[p] = '(';
    long long tdp = DP();
    if (tdp < k) {
      aa[p] = ')';
      k -= tdp;
    }
  }
  for (int i = 1; i <= n; i++, puts(""))
    for (int j = 1; j <= m; j++) putchar(aa[i + j - 1]);
}
int main() {
  build();
  while (__--) {
    init();
    solve();
  }
}
