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;
  struct Dsu {
    int n, m;
    vector<vector<pair<int, int>>> par;
    Dsu(int n_ = 0, int m_ = 0) : n(n_), m(m_) {
      par.assign(n, vector<pair<int, int>>(m));
    }
    bool valid(int x, int y) { return min(x, y) >= 0 && min(n - x, m - y) > 0; }
    pair<int, int> pull(int u, int v, const int x) {
      auto [p, q] = par[u][v];
      if ((u == p && v == q) || q != x) return {u, v};
      return par[u][v] = pull(p, q, x);
    }
    pair<int, int> get(int u, int v) {
      auto [p, q] = par[u][v];
      assert(valid(u, v) && valid(p, q));
      if (u == p && v == q) return {u, v};
      if (u == p) {
        par[u][v] = valid(u + 1, v) ? make_pair(u + 1, v) : make_pair(u, v);
      } else {
        assert(v == q);
        pull(u, v, v);
      }
      return get(p, q);
    }
  };
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m, k;
    cin >> n >> m >> k;
    Dsu dsu(n, m);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        int x, u = i, v = j;
        cin >> x;
        if (x == 1) v++;
        if (x == 2) u++;
        if (x == 3) v--;
        dsu.par[i][j] = dsu.valid(u, v) ? make_pair(u, v) : make_pair(i, j);
      }
    }
    for (int i = 0; i < k; i++) {
      int x;
      cin >> x;
      int k = dsu.get(0, --x).second;
      cout << k + 1 << ' ';
    }
    cout << '\n';
    return 0;
  }
  