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>
  long long int inf = 1e18;
  using namespace std;
  long long int add(long long a, long long b) {
    return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
  }
  long long int subtract(long long a, long long b) {
    return ((a % 1000000007) - (b % 1000000007)) % 1000000007;
  }
  long long int mult(long long a, long long b) {
    return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  }
  long long int add1(long long a, long long b) {
    return ((a % 998244353) + (b % 998244353)) % 998244353;
  }
  long long int subtract1(long long a, long long b) {
    return ((a % 998244353) - (b % 998244353)) % 998244353;
  }
  long long int mult1(long long a, long long b) {
    return ((a % 998244353) * (b % 998244353)) % 998244353;
  }
  long long int binmodwithMOD1(long long int a, long long int b) {
    long long int res = 1;
    while (b) {
      if (b & 1) res = mult1(res, a) % 998244353;
      a = mult1(a, a) % 998244353;
      b >>= 1;
    }
    return res;
  }
  long long int binmodwithMOD(long long int a, long long int b) {
    long long int res = 1;
    while (b) {
      if (b & 1) res = mult(res, a) % 1000000007;
      a = mult(a, a) % 1000000007;
      b >>= 1;
    }
    return res;
  }
  long long int binmodwithoutMOD1orMOD2(long long int a, long long int b) {
    long long int res = 1;
    while (b) {
      if (b & 1) res = res * a;
      a = a * a;
      b >>= 1;
    }
    return res;
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long int inverse(long long int n) {
    return binmodwithMOD(n, 1000000007 - 2);
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int n, m, k;
    cin >> n >> m >> k;
    vector<vector<long long int> > a(n, vector<long long int>(m)),
        tmp(n, vector<long long int>(m, -1));
    for (long long int i = 0; i < n; i++) {
      for (long long int j = 0; j < m; j++) cin >> a[i][j];
    }
    map<pair<long long int, long long int>, long long int> mp;
    vector<long long int> ans(k, 0);
    vector<long long int> balls(k, 0);
    for (long long int i = 0; i < k; i++) cin >> balls[i];
    long long int cnt = 1001;
    set<pair<long long int, long long int> > s1;
    for (long long int x = 0; x < k; x++) {
      long long int i = 0;
      long long int j = balls[x];
      j--;
      while (i < n) {
        if (a[i][j] == 2)
          i++;
        else {
          if (a[i][j] == 1)
            a[i][j++] = 2;
          else
            a[i][j--] = 2;
        }
      }
      cout << j + 1 << " ";
    }
    cout << "\n";
    cout << "\n";
    return 0;
  }
  