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;
  using ll = long long;
  using LL = __int128;
  template <class T>
  class BIT {
   private:
    int N;
    vector<T> nodes;
  
   public:
    BIT(int N) {
      this->N = N;
      nodes.assign(N + 1, 0);
    }
    void add(int k, T x) {
      for (int i = k; i <= N; i += (i & -i)) {
        nodes[i] += x;
      }
    }
    T sum(int k) {
      T result = 0;
      for (int i = k - 1; i > 0; i -= (i & -i)) {
        result += nodes[i];
      }
      return result;
    }
    T sum(int s, int t) { return sum(t) - sum(s); }
    int lower_bound(T x) {
      if (sum(N + 1) < x) return -1;
      int ok = N + 1;
      int ng = 0;
      while (ok - ng > 1) {
        int trying = (ok + ng) / 2;
        if (sum(trying + 1) >= x) {
          ok = trying;
        } else {
          ng = trying;
        }
      }
      return ok;
    }
    int upper_bound(T x) {
      if (sum(N + 1) <= x) return -1;
      int ok = N + 1;
      int ng = 0;
      while (ok - ng > 1) {
        int trying = (ok + ng) / 2;
        if (sum(trying + 1) > x) {
          ok = trying;
        } else {
          ng = trying;
        }
      }
      return ok;
    }
  };
  int n, m, k;
  int a[1000][1000];
  vector<BIT<int>> bit;
  bool is_ok(int l, int k, int c) { return bit[c].sum(l, k) == 0; }
  ll find_max_ok(ll lower, ll upper, int c) {
    ll ok = lower;
    ll ng = upper;
    while (ng - ok > 1) {
      ll trying = (ng + ok) / 2;
      if (is_ok(lower, trying, c)) {
        ok = trying;
      } else {
        ng = trying;
      }
    }
    return ok;
  }
  void solve() {
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++) {
      bit.push_back(BIT<int>(n));
    }
    for (int r = 0; r < n; r++) {
      for (int c = 0; c < m; c++) {
        cin >> a[r][c];
        if (a[r][c] != 2) bit[c].add(r + 1, 1);
      }
    }
    for (int t = 0; t < k; t++) {
      int c;
      cin >> c;
      c--;
      int r = 0;
      while (1) {
        r = find_max_ok(r + 1, n + 2, c) - 1;
        if (r == n) {
          cout << c + 1 << " ";
          break;
        }
        if (a[r][c] == 1) {
          a[r][c] = 2;
          bit[c].add(r + 1, -1);
          c++;
        } else if (a[r][c] == 3) {
          a[r][c] = 2;
          bit[c].add(r + 1, -1);
          c--;
        }
      }
    }
    cout << endl;
  }
  int main() {
    cin.tie(0), cout.tie(0), ios_base::sync_with_stdio(0);
    cout << fixed << setprecision(10);
    solve();
    return 0;
  }
  