#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
int solve(int x, int y, vector<vector<int>>& jump) {
  if (x == jump.size()) return x;
  if (jump[x][y] != -1) {
    int j = solve(jump[x][y], y, jump);
    jump[x][y] = j;
    return j;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j, k, n, m, x, y;
  cin >> n >> m >> k;
  vector<vector<int>> mat(n, vector<int>(m)), jump(n, vector<int>(m, -1));
  vector<int> col(k);
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j) cin >> mat[i][j];
  for (i = 0; i < k; ++i) cin >> col[i], --col[i];
  for (i = 0; i < n; ++i) {
    for (j = 0; j < m; ++j) {
      if (mat[i][j] == 2 && jump[i][j] == -1) {
        for (y = i; y < n; ++y)
          if (mat[y][j] != 2) break;
        x = y;
        for (y = i; y < n; ++y) {
          if (mat[y][j] == 2)
            jump[y][j] = x;
          else
            break;
        }
      }
    }
  }
  for (i = 0; i < k; ++i) {
    x = 0;
    y = col[i];
    while (true) {
      if (x == n) break;
      if (mat[x][y] == 1) {
        mat[x][y] = 2;
        if (x + 1 < n && jump[x + 1][y] != -1)
          jump[x][y] = jump[x + 1][y];
        else
          jump[x][y] = x + 1;
        y++;
      } else if (mat[x][y] == 3) {
        mat[x][y] = 2;
        if (x + 1 < n && jump[x + 1][y] != -1)
          jump[x][y] = jump[x + 1][y];
        else
          jump[x][y] = x + 1;
        y--;
      } else {
        x = solve(x, y, jump);
      }
    }
    cout << y + 1 << " ";
  }
  cout << "\n";
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
int solve(int x, int y, vector<vector<int>>& jump) {
  if (x == jump.size()) return x;
  if (jump[x][y] != -1) {
    int j = solve(jump[x][y], y, jump);
    jump[x][y] = j;
    return j;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j, k, n, m, x, y;
  cin >> n >> m >> k;
  vector<vector<int>> mat(n, vector<int>(m)), jump(n, vector<int>(m, -1));
  vector<int> col(k);
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j) cin >> mat[i][j];
  for (i = 0; i < k; ++i) cin >> col[i], --col[i];
  for (i = 0; i < n; ++i) {
    for (j = 0; j < m; ++j) {
      if (mat[i][j] == 2 && jump[i][j] == -1) {
        for (y = i; y < n; ++y)
          if (mat[y][j] != 2) break;
        x = y;
        for (y = i; y < n; ++y) {
          if (mat[y][j] == 2)
            jump[y][j] = x;
          else
            break;
        }
      }
    }
  }
  for (i = 0; i < k; ++i) {
    x = 0;
    y = col[i];
    while (true) {
      if (x == n) break;
      if (mat[x][y] == 1) {
        mat[x][y] = 2;
        if (x + 1 < n && jump[x + 1][y] != -1)
          jump[x][y] = jump[x + 1][y];
        else
          jump[x][y] = x + 1;
        y++;
      } else if (mat[x][y] == 3) {
        mat[x][y] = 2;
        if (x + 1 < n && jump[x + 1][y] != -1)
          jump[x][y] = jump[x + 1][y];
        else
          jump[x][y] = x + 1;
        y--;
      } else {
        x = solve(x, y, jump);
      }
    }
    cout << y + 1 << " ";
  }
  cout << "\n";
  return 0;
}

```