#include <bits/stdc++.h>
using namespace std;
long long solve();
vector<long long> a;
string s;
long long n;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
long long in(vector<long long> &a, long long n) {
  long long i = 0, x, sum = 0;
  for (i = 0; i < n; i++) {
    cin >> x;
    a.push_back({x});
  }
  return sum;
}
long long cnt[1001][1001];
long long mat[1001][1001];
long long m;
long long find(long long row, long long curcolumn) {
  if (row == n + 1) return curcolumn;
  long long ans;
  if (mat[row][curcolumn] == 1) {
    mat[row][curcolumn] = 2;
    ans = find(row, curcolumn + 1);
    if (row < n)
      cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
    else
      cnt[row][curcolumn] = 1;
    return ans;
  }
  if (mat[row][curcolumn] == 3) {
    mat[row][curcolumn] = 2;
    ans = find(row, curcolumn - 1);
    if (row < n)
      cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
    else
      cnt[row][curcolumn] = 1;
    return ans;
  }
  if (cnt[row][curcolumn]) {
    ans = find(row + cnt[row][curcolumn], curcolumn);
  } else
    ans = find(row + 1, curcolumn);
  if (row < n)
    cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
  else
    cnt[row][curcolumn] = 1;
  return ans;
}
long long solve() {
  a.clear();
  s.clear();
  long long i = 0, j = 0, k = 0;
  long long flag = 0;
  long long x = 0, y = 0, z = 0;
  cin >> n;
  cin >> m >> k;
  memset(cnt, 0, sizeof(cnt));
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      cin >> mat[i][j];
    }
  }
  long long col;
  for (i = 0; i < k; i++) {
    cin >> col;
    cout << find(1, col) << "\n";
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long solve();
vector<long long> a;
string s;
long long n;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
long long in(vector<long long> &a, long long n) {
  long long i = 0, x, sum = 0;
  for (i = 0; i < n; i++) {
    cin >> x;
    a.push_back({x});
  }
  return sum;
}
long long cnt[1001][1001];
long long mat[1001][1001];
long long m;
long long find(long long row, long long curcolumn) {
  if (row == n + 1) return curcolumn;
  long long ans;
  if (mat[row][curcolumn] == 1) {
    mat[row][curcolumn] = 2;
    ans = find(row, curcolumn + 1);
    if (row < n)
      cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
    else
      cnt[row][curcolumn] = 1;
    return ans;
  }
  if (mat[row][curcolumn] == 3) {
    mat[row][curcolumn] = 2;
    ans = find(row, curcolumn - 1);
    if (row < n)
      cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
    else
      cnt[row][curcolumn] = 1;
    return ans;
  }
  if (cnt[row][curcolumn]) {
    ans = find(row + cnt[row][curcolumn], curcolumn);
  } else
    ans = find(row + 1, curcolumn);
  if (row < n)
    cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
  else
    cnt[row][curcolumn] = 1;
  return ans;
}
long long solve() {
  a.clear();
  s.clear();
  long long i = 0, j = 0, k = 0;
  long long flag = 0;
  long long x = 0, y = 0, z = 0;
  cin >> n;
  cin >> m >> k;
  memset(cnt, 0, sizeof(cnt));
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      cin >> mat[i][j];
    }
  }
  long long col;
  for (i = 0; i < k; i++) {
    cin >> col;
    cout << find(1, col) << "\n";
  }
  return 0;
}

```