#include <bits/stdc++.h>
using namespace std;
int dx[]{-1, 0, 1, 0, 1, 1, -1, -1};
int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
long long const mod = 1e9 + 7;
int const N = 1e5 + 10;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  int g[n][m];
  vector<set<int>> s(m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> g[i][j];
      if (g[i][j] != 2) {
        s[j].insert(i);
      }
    }
  }
  for (int i = 0; i < k; i++) {
    int c, r = 0;
    cin >> c;
    --c;
    while (1) {
      if (g[r][c] == 1) {
        g[r][c] = 2;
        ++c;
      } else if (g[r][c] == 3) {
        g[r][c] = 2;
        --c;
      } else
        r++;
      if (r == n) {
        cout << c + 1 << " ";
        break;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int dx[]{-1, 0, 1, 0, 1, 1, -1, -1};
int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
long long const mod = 1e9 + 7;
int const N = 1e5 + 10;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  int g[n][m];
  vector<set<int>> s(m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> g[i][j];
      if (g[i][j] != 2) {
        s[j].insert(i);
      }
    }
  }
  for (int i = 0; i < k; i++) {
    int c, r = 0;
    cin >> c;
    --c;
    while (1) {
      if (g[r][c] == 1) {
        g[r][c] = 2;
        ++c;
      } else if (g[r][c] == 3) {
        g[r][c] = 2;
        --c;
      } else
        r++;
      if (r == n) {
        cout << c + 1 << " ";
        break;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```