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>
  void _print(T x) {
    cerr << x << endl;
  }
  template <class S, class T>
  void _print(pair<S, T> x) {
    cerr << "{" << x.first << "," << x.second << "}" << endl;
  }
  template <class T>
  void _print(vector<T> x) {
    cerr << "[";
    for (auto i : x) cerr << i << ", ";
    cerr << "]\n";
  }
  template <class S, class T>
  void _print(vector<pair<S, T>> x) {
    for (auto i : x) cerr << "[" << i.first << "," << i.second << "]" << endl;
  }
  template <class T>
  void _print(set<T> x) {
    cerr << "{ ";
    for (auto i : x) cerr << i << ", ";
    cerr << "}\n";
  }
  template <class S, class T>
  void _print(map<S, T> x) {
    for (auto i : x) {
      cerr << i.first << ": " << i.second << endl;
    }
  }
  long long exp(int a, int b, int p) {
    long long res = 1;
    a = a % p;
    while (b) {
      if (b & 1) res = (res * a) % p;
      a = ((long long)a * a) % p;
      b = b >> 1;
    }
    return res;
  }
  long long mmiPrime(int a, int p) { return exp(a, p - 2, p); }
  pair<int, int> extendedGCD(int a, int b) {
    if (b == 0) {
      pair<int, int> ans;
      ans.first = 1;
      ans.second = 0;
      return ans;
    }
    pair<int, int> temp = extendedGCD(b, a % b);
    pair<int, int> ans;
    ans.first = temp.second;
    ans.second = temp.first - (a / b) * temp.second;
    return ans;
  }
  int mmi(int a, int m) {
    pair<int, int> tmp = extendedGCD(a, m);
    return tmp.first;
  }
  int ncr(int n, int r, int p, int fact[]) {
    return ((fact[n] * mmiPrime(fact[r], p) % p) * mmiPrime(fact[n - r], p)) % p;
  }
  int n, m, k;
  int **mat;
  void solve() {
    cin >> n >> m >> k;
    mat = new int *[n];
    for (int i = 0; i < n; i++) {
      mat[i] = new int[m];
      for (int j = 0; j < m; j++) cin >> mat[i][j];
    }
    int c;
    int ans[k];
    for (int i = 0; i < k; i++) {
      cin >> c;
      --c;
      int row = 0;
      while (row < n) {
        if (mat[row][c] == 2)
          row++;
        else if (mat[row][c] == 1) {
          mat[row][c] = 2;
          c++;
        } else {
          mat[row][c] = 2;
          c--;
        }
      }
      ans[i] = c + 1;
    }
    for (int i : ans) cout << i << " ";
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    clock_t time = clock();
    solve();
    time = clock() - time;
    return 0;
  }
  