#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
template <class T>
T sqr(T x) {
  return x * x;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T> v_) {
  os << "{";
  for (long long i = 0; i < v_.size(); i++) {
    os << v_[i];
    if (i != v_.size() - 1) {
      os << ", ";
    }
  }
  os << "}";
  return os;
}
class DSU {
 private:
  const long long reserve = 5;

 public:
  vector<long long> parent;
  vector<long long> sets_size;
  explicit DSU(long long size) {
    parent = vector<long long>(size);
    sets_size = vector<long long>(size, 1);
    iota(parent.begin(), parent.end(), 0);
  }
  void make_set(long long v) {
    if (v >= parent.size()) {
      parent.resize(v + reserve);
      sets_size.resize(v + reserve);
    }
    parent[v] = v;
    sets_size[v] = 1;
  }
  long long find_set(long long v) {
    if (v == parent[v]) {
      return v;
    }
    return parent[v] = find_set(parent[v]);
  }
  void union_sets(long long a, long long b) {
    a = find_set(a);
    b = find_set(b);
    if (a != b) {
      parent[b] = a;
      sets_size[b] += sets_size[a];
    }
  }
  long long operator[](const long long& v) { return find_set(v); }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<long long>> a(n, vector<long long>(m, 0));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      cin >> a[i][j];
    }
  }
  DSU dsu(int32_t(n * m));
  for (long long q = 0; q < k; q++) {
    long long pos;
    cin >> pos;
    pos--;
    for (long long i = 0; i < n;) {
      while (a[i][pos] != 2) {
        long long to = a[i][pos] == 1 ? pos + 1 : pos - 1;
        a[i][pos] = 2;
        pos = to;
      }
      if (a[i][pos] == 2) {
        if (i + 1 == n) {
          break;
        }
        if (a[i + 1][pos] == 2) {
          dsu.union_sets(m * (i + 1) + pos, m * i + pos);
          i = dsu.find_set(m * i + pos) / m;
        } else {
          i++;
        }
      }
    }
    cout << pos + 1 << " ";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
template <class T>
T sqr(T x) {
  return x * x;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T> v_) {
  os << "{";
  for (long long i = 0; i < v_.size(); i++) {
    os << v_[i];
    if (i != v_.size() - 1) {
      os << ", ";
    }
  }
  os << "}";
  return os;
}
class DSU {
 private:
  const long long reserve = 5;

 public:
  vector<long long> parent;
  vector<long long> sets_size;
  explicit DSU(long long size) {
    parent = vector<long long>(size);
    sets_size = vector<long long>(size, 1);
    iota(parent.begin(), parent.end(), 0);
  }
  void make_set(long long v) {
    if (v >= parent.size()) {
      parent.resize(v + reserve);
      sets_size.resize(v + reserve);
    }
    parent[v] = v;
    sets_size[v] = 1;
  }
  long long find_set(long long v) {
    if (v == parent[v]) {
      return v;
    }
    return parent[v] = find_set(parent[v]);
  }
  void union_sets(long long a, long long b) {
    a = find_set(a);
    b = find_set(b);
    if (a != b) {
      parent[b] = a;
      sets_size[b] += sets_size[a];
    }
  }
  long long operator[](const long long& v) { return find_set(v); }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<long long>> a(n, vector<long long>(m, 0));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      cin >> a[i][j];
    }
  }
  DSU dsu(int32_t(n * m));
  for (long long q = 0; q < k; q++) {
    long long pos;
    cin >> pos;
    pos--;
    for (long long i = 0; i < n;) {
      while (a[i][pos] != 2) {
        long long to = a[i][pos] == 1 ? pos + 1 : pos - 1;
        a[i][pos] = 2;
        pos = to;
      }
      if (a[i][pos] == 2) {
        if (i + 1 == n) {
          break;
        }
        if (a[i + 1][pos] == 2) {
          dsu.union_sets(m * (i + 1) + pos, m * i + pos);
          i = dsu.find_set(m * i + pos) / m;
        } else {
          i++;
        }
      }
    }
    cout << pos + 1 << " ";
  }
  return 0;
}

```