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;
  char _buf[100000], *_p1 = _buf, *_p2 = _buf;
  inline int gi() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 3) + (x << 1) + (ch ^ 48);
      ch = getchar();
    }
    return (f == 1) ? x : -x;
  }
  const int maxn = 1005, maxm = 1e5 + 5;
  int a[maxn][maxn], n, m, k;
  int c[maxm];
  int fa[maxn][maxn];
  inline int get(int *fa, int x) {
    if (x == fa[x]) return x;
    return fa[x] = get(fa, fa[x]);
  }
  inline void init() {}
  inline void input() {
    n = gi(), m = gi(), k = gi();
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) a[i][j] = gi();
  }
  inline int go(int y) {
    int x = 1;
    while (x <= n) {
      x = get(fa[y], x);
      if (x != n) fa[y][x] = get(fa[y], fa[y][x + 1]);
      if (a[x][y] == 1)
        a[x][y] = 2, y++;
      else if (a[x][y] == 2)
        a[x][y] = 2, x++;
      else if (a[x][y] == 3)
        a[x][y] = 2, y--;
    }
    return y;
  }
  inline void solve() {
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) fa[j][i] = i;
    for (int i = 1; i <= k; ++i) {
      c[i] = gi();
      printf("%d ", go(c[i]));
    }
  }
  int main() {
    int T = 1;
    while (T--) {
      init();
      input();
      solve();
    }
    return 0;
  }
  