#include <bits/stdc++.h>
using namespace std;
const int NMAX = 1000;
int N, M, K;
int a[NMAX + 2][NMAX + 2];
struct state {
  int dad, sz;
  int down;
};
state ds[NMAX + 2][NMAX + 2];
bool active[NMAX + 2][NMAX + 2];
int root(int col, int p) {
  if (ds[p][col].dad != p) {
    return ds[p][col].dad = root(col, ds[p][col].dad);
  }
  return p;
}
void join(int col, int p, int q) {
  p = root(col, p);
  q = root(col, q);
  if (ds[p][col].sz > ds[q][col].sz) {
    swap(p, q);
  }
  ds[p][col].dad = q;
  ds[q][col].sz += ds[p][col].sz;
  ds[q][col].down = max(ds[q][col].down, ds[p][col].down);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> M >> K;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) {
      cin >> a[i][j];
      ds[i][j].dad = i;
      ds[i][j].sz = 1;
      ds[i][j].down = i;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) {
      if (a[i][j] == 2) {
        active[i][j] = true;
        if (active[i - 1][j] == true) {
          join(j, i - 1, i);
        }
      }
    }
  }
  for (int i = 1; i <= K; i++) {
    int x = 1, y;
    cin >> y;
    while (true) {
      if (active[x][y] == true) {
        int r = root(y, x);
        x = ds[r][y].down + 1;
      }
      if (x >= N + 1) {
        cout << y << ' ';
        break;
      }
      int old_x = x, old_y = y;
      if (a[x][y] == 1) {
        y++;
      } else if (a[x][y] == 2) {
        x++;
      } else {
        y--;
      }
      a[old_x][old_y] = 2;
      active[old_x][old_y] = true;
      if (active[old_x - 1][old_y] == true) {
        join(old_y, old_x - 1, old_x);
      }
      if (active[old_x + 1][old_y] == true) {
        join(old_y, old_x + 1, old_x);
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
const int NMAX = 1000;
int N, M, K;
int a[NMAX + 2][NMAX + 2];
struct state {
  int dad, sz;
  int down;
};
state ds[NMAX + 2][NMAX + 2];
bool active[NMAX + 2][NMAX + 2];
int root(int col, int p) {
  if (ds[p][col].dad != p) {
    return ds[p][col].dad = root(col, ds[p][col].dad);
  }
  return p;
}
void join(int col, int p, int q) {
  p = root(col, p);
  q = root(col, q);
  if (ds[p][col].sz > ds[q][col].sz) {
    swap(p, q);
  }
  ds[p][col].dad = q;
  ds[q][col].sz += ds[p][col].sz;
  ds[q][col].down = max(ds[q][col].down, ds[p][col].down);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> M >> K;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) {
      cin >> a[i][j];
      ds[i][j].dad = i;
      ds[i][j].sz = 1;
      ds[i][j].down = i;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) {
      if (a[i][j] == 2) {
        active[i][j] = true;
        if (active[i - 1][j] == true) {
          join(j, i - 1, i);
        }
      }
    }
  }
  for (int i = 1; i <= K; i++) {
    int x = 1, y;
    cin >> y;
    while (true) {
      if (active[x][y] == true) {
        int r = root(y, x);
        x = ds[r][y].down + 1;
      }
      if (x >= N + 1) {
        cout << y << ' ';
        break;
      }
      int old_x = x, old_y = y;
      if (a[x][y] == 1) {
        y++;
      } else if (a[x][y] == 2) {
        x++;
      } else {
        y--;
      }
      a[old_x][old_y] = 2;
      active[old_x][old_y] = true;
      if (active[old_x - 1][old_y] == true) {
        join(old_y, old_x - 1, old_x);
      }
      if (active[old_x + 1][old_y] == true) {
        join(old_y, old_x + 1, old_x);
      }
    }
  }
  return 0;
}

```