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;
  long long solve();
  vector<long long> a;
  string s;
  long long n;
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  long long in(vector<long long> &a, long long n) {
    long long i = 0, x, sum = 0;
    for (i = 0; i < n; i++) {
      cin >> x;
      a.push_back({x});
    }
    return sum;
  }
  long long cnt[1001][1001];
  long long mat[1001][1001];
  long long m;
  long long find(long long row, long long curcolumn) {
    if (row == n + 1) return curcolumn;
    long long ans;
    if (mat[row][curcolumn] == 1) {
      mat[row][curcolumn] = 2;
      ans = find(row, curcolumn + 1);
      if (row < n)
        cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
      else
        cnt[row][curcolumn] = 1;
      return ans;
    }
    if (mat[row][curcolumn] == 3) {
      mat[row][curcolumn] = 2;
      ans = find(row, curcolumn - 1);
      if (row < n)
        cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
      else
        cnt[row][curcolumn] = 1;
      return ans;
    }
    if (cnt[row][curcolumn]) {
      ans = find(row + cnt[row][curcolumn], curcolumn);
    } else
      ans = find(row + 1, curcolumn);
    if (row < n)
      cnt[row][curcolumn] = 1 + cnt[row + 1][curcolumn];
    else
      cnt[row][curcolumn] = 1;
    return ans;
  }
  long long solve() {
    a.clear();
    s.clear();
    long long i = 0, j = 0, k = 0;
    long long flag = 0;
    long long x = 0, y = 0, z = 0;
    cin >> n;
    cin >> m >> k;
    memset(cnt, 0, sizeof(cnt));
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        cin >> mat[i][j];
      }
    }
    long long col;
    for (i = 0; i < k; i++) {
      cin >> col;
      cout << find(1, col) << "\n";
    }
    return 0;
  }
  