#include <bits/stdc++.h>
using namespace std;
template <class T>
istream& operator>>(istream& in, vector<T>& vec) {
  for (auto& v : vec) {
    in >> v;
  }
  return in;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vec) {
  o << "[";
  for (const auto& v : vec) {
    o << v << ' ';
  }
  o << "]\n";
  return o;
}
int main() {
  int n, m, t;
  cin >> n >> m >> t;
  vector<vector<int>> mat(n, vector<int>(m));
  vector<vector<int>> bsm(m);
  cin >> mat;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (mat[i][j] != 2) {
        bsm[j].push_back(i);
      }
    }
  }
  while (t--) {
    int colI, rowI = -1;
    cin >> colI;
    colI--;
    while (true) {
      int newRowI = n;
      {
        auto& bsv = bsm[colI];
        int l = 0, r = bsv.size(), m;
        while (l < r) {
          m = (l + r) / 2;
          if (bsv[m] < rowI) {
            l = m + 1;
          } else {
            r = m;
          }
        }
        if (l < bsv.size()) {
          newRowI = bsv[l];
          int i = l + 1;
          for (; i < bsv.size() && bsv[i] == bsv[l]; i++) {
          }
          if (i < bsv.size()) {
            for (int a = l; a < i; a++) {
              bsv[a] = bsv[i];
            }
          } else {
            bsv.resize(bsv.size() - i + l);
          }
        }
      }
      rowI = newRowI;
      if (rowI >= n) {
        cout << colI + 1 << ' ';
        break;
      } else {
        if (mat[rowI][colI] == 1) {
          colI++;
        } else {
          colI--;
        }
      }
    }
  }
  cout << '\n';
}

### Prompt

In cpp, your task is to solve the following problem:

Mr. Chanek has a new game called Dropping Balls. Initially, Mr. Chanek has a grid a of size n × m

Each cell (x,y) contains an integer a_{x,y} denoting the direction of how the ball will move.

  * a_{x,y}=1 — the ball will move to the right (the next cell is (x, y + 1)); 
  * a_{x,y}=2 — the ball will move to the bottom (the next cell is (x + 1, y)); 
  * a_{x,y}=3 — the ball will move to the left (the next cell is (x, y - 1)). 



Every time a ball leaves a cell (x,y), the integer a_{x,y} will change to 2. Mr. Chanek will drop k balls sequentially, each starting from the first row, and on the c_1, c_2, ..., c_k-th (1 ≤ c_i ≤ m) columns.

Determine in which column each ball will end up in (position of the ball after leaving the grid).

Input

The first line contains three integers n, m, and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10^5) — the size of the grid and the number of balls dropped by Mr. Chanek.

The i-th of the next n lines contains m integers a_{i,1},a_{i,2},…,a_{i,m} (1 ≤ a_{i,j} ≤ 3). It will satisfy a_{i, 1} ≠ 3 and a_{i, m} ≠ 1.

The next line contains k integers c_1, c_2, …, c_k (1 ≤ c_i ≤ m) — the balls' column positions dropped by Mr. Chanek sequentially.

Output

Output k integers — the i-th integer denoting the column where the i-th ball will end.

Examples

Input


5 5 3
1 2 3 3 3
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
1 2 1


Output


2 2 1 


Input


1 2 2
1 3
1 2


Output


1 2 

Note

In the first example, the first ball will drop as follows. Note that the cell (1, 1) will change direction to the bottom direction.

<image>

The second and third balls will drop as follows. 

<image>

All balls will be dropped from the first row and on the c_1, c_2, ..., c_k-th columns respectively. A ball will stop dropping once it leaves the grid.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
istream& operator>>(istream& in, vector<T>& vec) {
  for (auto& v : vec) {
    in >> v;
  }
  return in;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vec) {
  o << "[";
  for (const auto& v : vec) {
    o << v << ' ';
  }
  o << "]\n";
  return o;
}
int main() {
  int n, m, t;
  cin >> n >> m >> t;
  vector<vector<int>> mat(n, vector<int>(m));
  vector<vector<int>> bsm(m);
  cin >> mat;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (mat[i][j] != 2) {
        bsm[j].push_back(i);
      }
    }
  }
  while (t--) {
    int colI, rowI = -1;
    cin >> colI;
    colI--;
    while (true) {
      int newRowI = n;
      {
        auto& bsv = bsm[colI];
        int l = 0, r = bsv.size(), m;
        while (l < r) {
          m = (l + r) / 2;
          if (bsv[m] < rowI) {
            l = m + 1;
          } else {
            r = m;
          }
        }
        if (l < bsv.size()) {
          newRowI = bsv[l];
          int i = l + 1;
          for (; i < bsv.size() && bsv[i] == bsv[l]; i++) {
          }
          if (i < bsv.size()) {
            for (int a = l; a < i; a++) {
              bsv[a] = bsv[i];
            }
          } else {
            bsv.resize(bsv.size() - i + l);
          }
        }
      }
      rowI = newRowI;
      if (rowI >= n) {
        cout << colI + 1 << ' ';
        break;
      } else {
        if (mat[rowI][colI] == 1) {
          colI++;
        } else {
          colI--;
        }
      }
    }
  }
  cout << '\n';
}

```