#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int N = 1010;
int n, m, k, top;
int a[N][N], c[1000050], ans[1000050];
pair<int, int> nxt[N][N];
int vis[N][N];
vector<pair<int, int>> upd;
void push(int row, int col) {
  for (auto [x, y] : upd) {
    nxt[x][y] = {row, col};
  }
  upd.clear();
}
void go(int i, int j) {
  if (i < 0 || i > n || j < 0 || j >= m) return;
  if (i == n) {
    ans[top++] = j + 1;
    return;
  }
  if (nxt[i][j].first != -1) {
    a[i][j] = 2;
    tie(i, j) = nxt[i][j];
    go(i, j);
    return;
  }
  if (a[i][j] == 1) {
    a[i][j] = 2;
    go(i, j + 1);
    nxt[i][j] = nxt[i + 1][j];
  } else if (a[i][j] == 2) {
    a[i][j] = 2;
    go(i + 1, j);
    nxt[i][j] = nxt[i + 1][j];
  } else {
    a[i][j] = 2;
    go(i, j - 1);
    nxt[i][j] = nxt[i + 1][j];
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  for (int i = 0; i < k; i++) {
    scanf("%d", &c[i]);
    --c[i];
    int x = 0, y = c[i];
    while (x < n) {
      int ox = x, oy = y;
      if (a[x][y] == 1)
        y++;
      else if (a[x][y] == 2)
        x++;
      else
        y--;
      a[ox][oy] = 2;
    }
    printf("%d ", y + 1);
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
using ld = long double;
const int N = 1010;
int n, m, k, top;
int a[N][N], c[1000050], ans[1000050];
pair<int, int> nxt[N][N];
int vis[N][N];
vector<pair<int, int>> upd;
void push(int row, int col) {
  for (auto [x, y] : upd) {
    nxt[x][y] = {row, col};
  }
  upd.clear();
}
void go(int i, int j) {
  if (i < 0 || i > n || j < 0 || j >= m) return;
  if (i == n) {
    ans[top++] = j + 1;
    return;
  }
  if (nxt[i][j].first != -1) {
    a[i][j] = 2;
    tie(i, j) = nxt[i][j];
    go(i, j);
    return;
  }
  if (a[i][j] == 1) {
    a[i][j] = 2;
    go(i, j + 1);
    nxt[i][j] = nxt[i + 1][j];
  } else if (a[i][j] == 2) {
    a[i][j] = 2;
    go(i + 1, j);
    nxt[i][j] = nxt[i + 1][j];
  } else {
    a[i][j] = 2;
    go(i, j - 1);
    nxt[i][j] = nxt[i + 1][j];
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  for (int i = 0; i < k; i++) {
    scanf("%d", &c[i]);
    --c[i];
    int x = 0, y = c[i];
    while (x < n) {
      int ox = x, oy = y;
      if (a[x][y] == 1)
        y++;
      else if (a[x][y] == 2)
        x++;
      else
        y--;
      a[ox][oy] = 2;
    }
    printf("%d ", y + 1);
  }
}

```