#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,O2,O3,unroll-loops")
#pragma GCC target("avx2")
using namespace std;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << "[" << H << "]";
  debug_out(T...);
}
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const double eps = 0.000001;
const int MOD = 998244353;
const int INF = 1000000101;
const long long LLINF = 1223372000000000555;
const int N = 2e6 + 3e2;
const int M = 5222;
int a[M][M];
int nxt[M][M];
void solve(int TC) {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i <= k; i++) {
    int x = 1, y;
    cin >> y;
    while (x <= n) {
      if (a[x][y] == 2) {
        x++;
        continue;
      }
      if (a[x][y] == 1) {
        int j = y;
        while (a[x][j] == 1) {
          a[x][j] = 2;
          j++;
        }
        if (a[x][j] == 3)
          y = j - 1;
        else
          y = j;
        a[x][j] = 2;
        x++;
        continue;
      }
      if (a[x][y] == 3) {
        int j = y;
        while (a[x][j] == 3) {
          a[x][j] = 2;
          j--;
        }
        if (a[x][j] == 1)
          y = j + 1;
        else
          y = j;
        a[x][j] = 2;
        x++;
      }
    }
    cout << y << ' ';
  }
}
int main() {
  startTime = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int TC = 1;
  for (int test = 1; test <= TC; test++) {
    solve(test);
  }
  cerr << endl << "Time: " << int(getCurrentTime() * 1000) << " ms" << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
#pragma GCC optimize("Ofast,O2,O3,unroll-loops")
#pragma GCC target("avx2")
using namespace std;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << "[" << H << "]";
  debug_out(T...);
}
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const double eps = 0.000001;
const int MOD = 998244353;
const int INF = 1000000101;
const long long LLINF = 1223372000000000555;
const int N = 2e6 + 3e2;
const int M = 5222;
int a[M][M];
int nxt[M][M];
void solve(int TC) {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 1; i <= k; i++) {
    int x = 1, y;
    cin >> y;
    while (x <= n) {
      if (a[x][y] == 2) {
        x++;
        continue;
      }
      if (a[x][y] == 1) {
        int j = y;
        while (a[x][j] == 1) {
          a[x][j] = 2;
          j++;
        }
        if (a[x][j] == 3)
          y = j - 1;
        else
          y = j;
        a[x][j] = 2;
        x++;
        continue;
      }
      if (a[x][y] == 3) {
        int j = y;
        while (a[x][j] == 3) {
          a[x][j] = 2;
          j--;
        }
        if (a[x][j] == 1)
          y = j + 1;
        else
          y = j;
        a[x][j] = 2;
        x++;
      }
    }
    cout << y << ' ';
  }
}
int main() {
  startTime = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int TC = 1;
  for (int test = 1; test <= TC; test++) {
    solve(test);
  }
  cerr << endl << "Time: " << int(getCurrentTime() * 1000) << " ms" << endl;
  return 0;
}

```