#include <bits/stdc++.h>
using namespace std;
char _buf[100000], *_p1 = _buf, *_p2 = _buf;
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return (f == 1) ? x : -x;
}
const int maxn = 1005, maxm = 1e5 + 5;
int a[maxn][maxn], n, m, k;
int c[maxm];
int fa[maxn][maxn];
inline int get(int *fa, int x) {
  if (x == fa[x]) return x;
  return fa[x] = get(fa, fa[x]);
}
inline void init() {}
inline void input() {
  n = gi(), m = gi(), k = gi();
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) a[i][j] = gi();
}
inline int go(int y) {
  int x = 1;
  while (x <= n) {
    x = get(fa[y], x);
    if (x != n) fa[y][x] = get(fa[y], fa[y][x + 1]);
    if (a[x][y] == 1)
      a[x][y] = 2, y++;
    else if (a[x][y] == 2)
      a[x][y] = 2, x++;
    else if (a[x][y] == 3)
      a[x][y] = 2, y--;
  }
  return y;
}
inline void solve() {
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) fa[j][i] = i;
  for (int i = 1; i <= k; ++i) {
    c[i] = gi();
    printf("%d ", go(c[i]));
  }
}
int main() {
  int T = 1;
  while (T--) {
    init();
    input();
    solve();
  }
  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;
char _buf[100000], *_p1 = _buf, *_p2 = _buf;
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return (f == 1) ? x : -x;
}
const int maxn = 1005, maxm = 1e5 + 5;
int a[maxn][maxn], n, m, k;
int c[maxm];
int fa[maxn][maxn];
inline int get(int *fa, int x) {
  if (x == fa[x]) return x;
  return fa[x] = get(fa, fa[x]);
}
inline void init() {}
inline void input() {
  n = gi(), m = gi(), k = gi();
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) a[i][j] = gi();
}
inline int go(int y) {
  int x = 1;
  while (x <= n) {
    x = get(fa[y], x);
    if (x != n) fa[y][x] = get(fa[y], fa[y][x + 1]);
    if (a[x][y] == 1)
      a[x][y] = 2, y++;
    else if (a[x][y] == 2)
      a[x][y] = 2, x++;
    else if (a[x][y] == 3)
      a[x][y] = 2, y--;
  }
  return y;
}
inline void solve() {
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) fa[j][i] = i;
  for (int i = 1; i <= k; ++i) {
    c[i] = gi();
    printf("%d ", go(c[i]));
  }
}
int main() {
  int T = 1;
  while (T--) {
    init();
    input();
    solve();
  }
  return 0;
}

```