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>
  using namespace std;
  template <class A, class B>
  ostream& operator<<(ostream& out, const pair<A, B>& p) {
    return out << "(" << p.first << ", " << p.second << ")";
  }
  template <class A>
  ostream& operator<<(ostream& out, const vector<A>& v) {
    for (int i = int(0); i < int(int((v).size())); i++) {
      if (i) out << " ";
      out << v[i];
    }
    return out;
  }
  const int N = 1005;
  const int mlen = 1000009;
  const long long maxl = 1001000000;
  const int md = 1000000007;
  vector<int> a;
  void fd() {}
  void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int> > v(n);
    vector<vector<bool> > isDrop(n + 1);
    for (int i = int(0); i < int(n); i++) {
      v[i].resize(m);
      isDrop[i].resize(m);
      for (int j = int(0); j < int(m); j++) cin >> v[i][j];
    }
    isDrop[n].resize(m);
    for (int j = int(0); j < int(m); j++) isDrop[n][j] = true;
    int x, startDrop;
    for (int q = int(0); q < int(k); q++) {
      cin >> x;
      x--;
      startDrop = -1;
      for (int i = int(0); i < int(n + 1); i++) {
        if (isDrop[i][x]) {
          if (startDrop != -1)
            for (int j = int(startDrop); j < int(n); j++) isDrop[j][x] = true;
          cout << ++x << ' ';
          break;
        }
        if (v[i][x] != 2) {
          startDrop = -1;
          int val = v[i][x];
          v[i][x] = 2;
          i--;
          if (val == 1)
            x++;
          else
            x--;
        } else if (startDrop == -1)
          startDrop = i;
      }
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) {
      solve();
      fflush(stdout);
    }
    return 0;
  }
  