#include <bits/stdc++.h>
using namespace std;
struct yqh {
  int x, y;
} fa[1020][1020];
int n, m, k;
int mp[1020][1020];
yqh getfa(int x, int y) {
  if (fa[x][y].x == x && fa[x][y].y == y) return fa[x][y];
  fa[x][y] = getfa(fa[x][y].x, fa[x][y].y);
  return fa[x][y];
}
int get() {
  char t = getchar();
  while (t < '0' || t > '9') t = getchar();
  int x = 0;
  while (t >= '0' && t <= '9') {
    x *= 10;
    x += t - '0';
    t = getchar();
  }
  return x;
}
int main() {
  n = get();
  m = get();
  k = get();
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      mp[i][j] = get();
      if (mp[i][j] == 2) {
        fa[i][j].x = i + 1;
        fa[i][j].y = j;
      } else {
        fa[i][j].x = i;
        fa[i][j].y = j;
      }
    }
  }
  for (int j = 1; j <= m; ++j) {
    fa[n + 1][j].x = n + 1;
    fa[n + 1][j].y = j;
  }
  for (int i = 1; i <= k; ++i) {
    int sx = 1;
    int sy = get();
    while (sx <= n) {
      yqh nex = getfa(sx, sy);
      sx = nex.x;
      sy = nex.y;
      if (sx <= n) {
        int flag = mp[sx][sy];
        mp[sx][sy] = 2;
        fa[sx][sy].x = sx + 1;
        fa[sx][sy].y = sy;
        if (flag == 1)
          ++sy;
        else if (flag == 2)
          ++sx;
        else
          --sy;
      }
    }
    printf("%d", sy);
    if (i < k)
      printf(" ");
    else
      printf("\n");
  }
  return 0;
}

### Prompt

Generate 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;
struct yqh {
  int x, y;
} fa[1020][1020];
int n, m, k;
int mp[1020][1020];
yqh getfa(int x, int y) {
  if (fa[x][y].x == x && fa[x][y].y == y) return fa[x][y];
  fa[x][y] = getfa(fa[x][y].x, fa[x][y].y);
  return fa[x][y];
}
int get() {
  char t = getchar();
  while (t < '0' || t > '9') t = getchar();
  int x = 0;
  while (t >= '0' && t <= '9') {
    x *= 10;
    x += t - '0';
    t = getchar();
  }
  return x;
}
int main() {
  n = get();
  m = get();
  k = get();
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      mp[i][j] = get();
      if (mp[i][j] == 2) {
        fa[i][j].x = i + 1;
        fa[i][j].y = j;
      } else {
        fa[i][j].x = i;
        fa[i][j].y = j;
      }
    }
  }
  for (int j = 1; j <= m; ++j) {
    fa[n + 1][j].x = n + 1;
    fa[n + 1][j].y = j;
  }
  for (int i = 1; i <= k; ++i) {
    int sx = 1;
    int sy = get();
    while (sx <= n) {
      yqh nex = getfa(sx, sy);
      sx = nex.x;
      sy = nex.y;
      if (sx <= n) {
        int flag = mp[sx][sy];
        mp[sx][sy] = 2;
        fa[sx][sy].x = sx + 1;
        fa[sx][sy].y = sy;
        if (flag == 1)
          ++sy;
        else if (flag == 2)
          ++sx;
        else
          --sy;
      }
    }
    printf("%d", sy);
    if (i < k)
      printf(" ");
    else
      printf("\n");
  }
  return 0;
}

```