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 T>
  istream& operator>>(istream& in, vector<T>& vec) {
    for (auto& v : vec) {
      in >> v;
    }
    return in;
  }
  template <class T>
  ostream& operator<<(ostream& o, const vector<T>& vec) {
    o << "[";
    for (const auto& v : vec) {
      o << v << ' ';
    }
    o << "]\n";
    return o;
  }
  int main() {
    int n, m, t;
    cin >> n >> m >> t;
    vector<vector<int>> mat(n, vector<int>(m));
    vector<vector<int>> bsm(m);
    cin >> mat;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (mat[i][j] != 2) {
          bsm[j].push_back(i);
        }
      }
    }
    while (t--) {
      int colI, rowI = -1;
      cin >> colI;
      colI--;
      while (true) {
        int newRowI = n;
        {
          auto& bsv = bsm[colI];
          int l = 0, r = bsv.size(), m;
          while (l < r) {
            m = (l + r) / 2;
            if (bsv[m] < rowI) {
              l = m + 1;
            } else {
              r = m;
            }
          }
          if (l < bsv.size()) {
            newRowI = bsv[l];
            int i = l + 1;
            for (; i < bsv.size() && bsv[i] == bsv[l]; i++) {
            }
            if (i < bsv.size()) {
              for (int a = l; a < i; a++) {
                bsv[a] = bsv[i];
              }
            } else {
              while (i-- > l) {
                bsv.pop_back();
              }
            }
          }
        }
        rowI = newRowI;
        if (rowI >= n) {
          cout << colI + 1 << ' ';
          break;
        } else {
          if (mat[rowI][colI] == 1) {
            colI++;
          } else {
            colI--;
          }
        }
      }
    }
    cout << '\n';
  }
  