#include <bits/stdc++.h>
using namespace std;
const long long int MOD = ((1e9) + 7);
const long long int INF = 1e18;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, k;
  cin >> n >> m >> k;
  vector<vector<long long int>> a(n + 2, (vector<long long int>(m + 2, 0)));
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j < m; j++) {
      if (a[i][j] == 1 && a[i][j + 1] == 3) {
        a[i][j] = 2;
        a[i][j + 1] = 2;
      }
    }
  }
  while (k--) {
    long long int x;
    cin >> x;
    long long int p1 = 1;
    long long int p2 = x;
    while (a[p1][p2] != 0 && p2 > 0 && p2 <= m) {
      if (a[p1][p2] == 2)
        p1++;
      else if (a[p1][p2] == 1) {
        a[p1][p2] = 2;
        p2++;
      } else {
        a[p1][p2] = 2;
        p2--;
      }
    }
    cout << p2 << " ";
  }
  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;
const long long int MOD = ((1e9) + 7);
const long long int INF = 1e18;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, k;
  cin >> n >> m >> k;
  vector<vector<long long int>> a(n + 2, (vector<long long int>(m + 2, 0)));
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j < m; j++) {
      if (a[i][j] == 1 && a[i][j + 1] == 3) {
        a[i][j] = 2;
        a[i][j + 1] = 2;
      }
    }
  }
  while (k--) {
    long long int x;
    cin >> x;
    long long int p1 = 1;
    long long int p2 = x;
    while (a[p1][p2] != 0 && p2 > 0 && p2 <= m) {
      if (a[p1][p2] == 2)
        p1++;
      else if (a[p1][p2] == 1) {
        a[p1][p2] = 2;
        p2++;
      } else {
        a[p1][p2] = 2;
        p2--;
      }
    }
    cout << p2 << " ";
  }
  return 0;
}

```