description: 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.
solution: #include <bits/stdc++.h>
  const int N = 100001, M = 998244353, mod = 1000000007;
  const long long MX = INT64_MAX, MN = INT64_MIN, oo = 1e18;
  using namespace std;
  vector<string> tokenizer(string str, char ch) {
    std::istringstream var((str));
    vector<string> v;
    string t;
    while (getline((var), t, (ch))) {
      v.emplace_back(t);
    }
    return v;
  }
  template <typename T1, typename T2>
  istream& operator>>(istream& in, pair<T1, T2>& a) {
    in >> a.first >> a.second;
    return in;
  }
  template <typename T1, typename T2>
  ostream& operator<<(ostream& out, pair<T1, T2> a) {
    out << a.first << " " << a.second;
    return out;
  }
  void err(istream_iterator<string> it) {}
  template <typename T, typename... Args>
  void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << "\n";
    err(++it, args...);
  }
  template <typename... T>
  void read(T&... args) {
    ((cin >> args), ...);
  }
  template <typename... T>
  void put(T&&... args) {
    ((cout << args << " "), ...);
  }
  void file_i_o() {
    freopen("./tests/test01.txt", "r", stdin);
    freopen("./tests/output01.txt", "w", stdout);
  }
  void solve(int T) {
    int n, m, k;
    read(n, m, k);
    vector<vector<int>> grid(n, vector<int>(m));
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) cin >> grid[i][j];
    while (k--) {
      int c;
      read(c);
      c--;
      int r = 0;
      while (r < n) {
        if (grid[r][c] == 2)
          r++;
        else {
          int nc = c;
          if (grid[r][c] == 1)
            nc++;
          else
            nc--;
          grid[r][c] = 2;
          c = nc;
        }
      }
      put(c + 1);
    }
  }
  int main(int argc, char const* argv[]) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    for (int i = 1; i < t + 1; i++) {
      solve(t);
    }
    return 0;
  }
  