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.
#include <bits/stdc++.h>
using namespace std;
int n, m, k;
int mp[1005][1005];
int cur;
int ji[1005];
int s;
char c;
int qr() {
  s = 0;
  c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s;
}
int main() {
  n = qr();
  m = qr();
  k = qr();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      mp[i][j] = qr();
    }
  }
  for (int i = 1; i <= m; i++) {
    ji[i] = n + 1;
    for (int j = n; j >= 1; j--) {
      if (mp[j][i] == 2) {
        ji[i] = j;
      } else {
        break;
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    cur = qr();
    for (int j = 1; j <= n; j++) {
      if (ji[cur] <= j) {
        break;
      }
      while (mp[j][cur] != 2) {
        if (mp[j][cur] == 1) {
          mp[j][cur] = 2;
          cur++;
        } else {
          mp[j][cur] = 2;
          cur--;
        }
      }
    }
    printf("%d ", cur);
    for (int j = ji[cur]; j >= 1; j--) {
      if (mp[j][cur] == 2) {
        ji[i] = j;
      } else {
        break;
      }
    }
  }
  return 0;
}
