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;
  const long long INF = 1e18;
  const long long MOD = 1e9 + 7;
  template <class T>
  T sqr(T x) {
    return x * x;
  }
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T> v_) {
    os << "{";
    for (long long i = 0; i < v_.size(); i++) {
      os << v_[i];
      if (i != v_.size() - 1) {
        os << ", ";
      }
    }
    os << "}";
    return os;
  }
  class DSU {
   private:
    const long long reserve = 5;
  
   public:
    vector<long long> parent;
    vector<long long> sets_size;
    explicit DSU(long long size) {
      parent = vector<long long>(size);
      sets_size = vector<long long>(size, 1);
      iota(parent.begin(), parent.end(), 0);
    }
    void make_set(long long v) {
      if (v >= parent.size()) {
        parent.resize(v + reserve);
        sets_size.resize(v + reserve);
      }
      parent[v] = v;
      sets_size[v] = 1;
    }
    long long find_set(long long v) {
      if (v == parent[v]) {
        return v;
      }
      return parent[v] = find_set(parent[v]);
    }
    void union_sets(long long a, long long b) {
      a = find_set(a);
      b = find_set(b);
      if (a != b) {
        parent[b] = a;
        sets_size[b] += sets_size[a];
      }
    }
    long long operator[](const long long& v) { return find_set(v); }
  };
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    long long n, m, k;
    cin >> n >> m >> k;
    vector<vector<long long>> a(n, vector<long long>(m, 0));
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < m; ++j) {
        cin >> a[i][j];
      }
    }
    DSU dsu(int32_t(n * m));
    for (long long q = 0; q < k; q++) {
      long long pos;
      cin >> pos;
      pos--;
      for (long long i = 0; i < n;) {
        while (a[i][pos] != 2) {
          long long to = a[i][pos] == 1 ? pos + 1 : pos - 1;
          a[i][pos] = 2;
          pos = to;
        }
        if (a[i][pos] == 2) {
          if (i + 1 == n) {
            break;
          }
          if (a[i + 1][pos] == 2) {
            dsu.union_sets(m * (i + 1) + pos, m * i + pos);
            i = dsu.find_set(m * i + pos) / m;
          } else {
            i++;
          }
        }
      }
      cout << pos + 1 << " ";
    }
    return 0;
  }
  