#include <bits/stdc++.h>
using namespace std;
int const N = 1e3 + 5;
class dsu {
 public:
  int _n;
  std::vector<int> p;
  dsu() : _n(0){};
  explicit dsu(int n) : _n(n), p(n, -1) {}
  int merge(int a, int b) {
    assert(0 <= a && a < _n);
    assert(0 <= b && b < _n);
    int x = leader(a), y = leader(b);
    if (x == y) return x;
    p[x] += p[y];
    p[y] = x;
    return x;
  }
  void init(int n) {
    p.resize(n);
    for (int i = 1; i < n; i++) p[i] = i;
  }
  bool same(int a, int b) {
    assert(0 <= a && a < _n);
    assert(0 <= b && b < _n);
    return leader(a) == leader(b);
  }
  int leader(int a) {
    assert(0 <= a && a < _n);
    if (p[a] < 0) return a;
    return (p[a] == a) ? a : p[a] = leader(p[a]);
  }
  int size(int a) {
    assert(0 <= a && a < _n);
    return -p[leader(a)];
  }
  std::vector<std::vector<int>> groups() {
    std::vector<int> leader_buf(_n), group_size(_n);
    for (int i = 0; i < _n; i++) {
      leader_buf[i] = leader(i);
      group_size[leader_buf[i]]++;
    }
    std::vector<std::vector<int>> result(_n);
    for (int i = 0; i < _n; i++) result[i].reserve(group_size[i]);
    for (int i = 0; i < _n; i++) result[leader_buf[i]].push_back(i);
    result.erase(
        std::remove_if(result.begin(), result.end(),
                       [&](const std::vector<int> &v) { return v.empty(); }),
        result.end());
    return result;
  }
};
int g[N][N];
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<dsu> v(m + 2, dsu(n + 2));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) cin >> g[i][j];
  }
  for (int i = 1; i <= m; i++) {
    v[i].init(n + 1);
    for (int j = 1; j <= n; j++) {
      if (g[j][i] == 2) v[i].p[j] = j + 1;
    }
  }
  while (k--) {
    int j;
    cin >> j;
    int i = v[j].leader(1);
    while (i <= n) {
      v[j].p[i] = i + 1;
      if (g[i][j] == 1)
        g[i][j++] = 2;
      else
        g[i][j--] = 2;
      i = v[j].leader(i);
      ;
    }
    cout << j << " ";
  }
}

### Prompt

In cpp, your task is to solve 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;
int const N = 1e3 + 5;
class dsu {
 public:
  int _n;
  std::vector<int> p;
  dsu() : _n(0){};
  explicit dsu(int n) : _n(n), p(n, -1) {}
  int merge(int a, int b) {
    assert(0 <= a && a < _n);
    assert(0 <= b && b < _n);
    int x = leader(a), y = leader(b);
    if (x == y) return x;
    p[x] += p[y];
    p[y] = x;
    return x;
  }
  void init(int n) {
    p.resize(n);
    for (int i = 1; i < n; i++) p[i] = i;
  }
  bool same(int a, int b) {
    assert(0 <= a && a < _n);
    assert(0 <= b && b < _n);
    return leader(a) == leader(b);
  }
  int leader(int a) {
    assert(0 <= a && a < _n);
    if (p[a] < 0) return a;
    return (p[a] == a) ? a : p[a] = leader(p[a]);
  }
  int size(int a) {
    assert(0 <= a && a < _n);
    return -p[leader(a)];
  }
  std::vector<std::vector<int>> groups() {
    std::vector<int> leader_buf(_n), group_size(_n);
    for (int i = 0; i < _n; i++) {
      leader_buf[i] = leader(i);
      group_size[leader_buf[i]]++;
    }
    std::vector<std::vector<int>> result(_n);
    for (int i = 0; i < _n; i++) result[i].reserve(group_size[i]);
    for (int i = 0; i < _n; i++) result[leader_buf[i]].push_back(i);
    result.erase(
        std::remove_if(result.begin(), result.end(),
                       [&](const std::vector<int> &v) { return v.empty(); }),
        result.end());
    return result;
  }
};
int g[N][N];
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<dsu> v(m + 2, dsu(n + 2));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) cin >> g[i][j];
  }
  for (int i = 1; i <= m; i++) {
    v[i].init(n + 1);
    for (int j = 1; j <= n; j++) {
      if (g[j][i] == 2) v[i].p[j] = j + 1;
    }
  }
  while (k--) {
    int j;
    cin >> j;
    int i = v[j].leader(1);
    while (i <= n) {
      v[j].p[i] = i + 1;
      if (g[i][j] == 1)
        g[i][j++] = 2;
      else
        g[i][j--] = 2;
      i = v[j].leader(i);
      ;
    }
    cout << j << " ";
  }
}

```