#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const long long inf = 1e15;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
int large_power(int x, unsigned int y) {
  int res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
int power(int x, int y) {
  int res = 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
string bin(int n) {
  string s = "";
  while (n > 1) {
    int rem = n % 2;
    if (rem)
      s = "1" + s;
    else
      s = "0" + s;
    n = n / 2;
  }
  if (n) s = "1" + s;
  return s;
}
long long exp(long long i, long long j) {
  if (j == 1) return i;
  if (j == 0) return 1;
  if (j & 1) {
    return (i * exp((i * i) % mod, j / 2)) % mod;
  }
  return exp((i * i) % mod, j / 2) % mod;
}
long long invMod(long long i) { return exp(i, mod - 2) % mod; }
long long nCr(long long n, long long r) {
  long long ans = 1;
  for (long long i = n; i > n - r; i--) {
    ans = (ans * i) % mod;
  }
  for (long long i = 1; i <= r; i++) {
    ans = (ans * invMod(i)) % mod;
  }
  return ans;
}
tuple<int, int> divisors(int n) {
  map<int, int> mp;
  int e = 0, o = 0;
  while (n % 2 == 0) {
    if (mp[2] > 0)
      mp[2]++;
    else
      mp[2] = 1;
    n /= 2;
    e += 1;
  }
  int i = 3;
  while (i * i <= n) {
    if (n % i == 0) {
      o += 1;
      if (mp[i] > 0)
        mp[i]++;
      else
        mp[i] = 1;
      n /= i;
    } else {
      i += 2;
    }
  }
  if (n > 2) {
    o += 1;
    if (mp[n] > 0)
      mp[n]++;
    else
      mp[n] = 1;
  }
  return make_tuple(e, o);
}
int n, m, k;
vector<int> ans;
int pi = -1, pj = -1;
void add(vector<vector<int>>& mat, int row, int col) {
  if (row == n) {
    ans.push_back(col + 1);
    pi = -1;
    pj = -1;
    return;
  }
  if (pi != -1 && pj != -1) {
    mat[pi][pj] = 2;
  }
  if (mat[row][col] == 2) {
    pi = row;
    pj = col;
    add(mat, row + 1, col);
    return;
  } else if (mat[row][col] == 1) {
    pi = row;
    pj = col;
    add(mat, row, col + 1);
    return;
  } else if (mat[row][col] == 3) {
    pi = row;
    pj = col;
    add(mat, row, col - 1);
    return;
  }
}
void solve() {
  cin >> n >> m >> k;
  vector<vector<int>> mat(n, vector<int>(m));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> mat[i][j];
    }
  }
  vector<int> balls(k);
  for (int i = 0; i < k; i++) {
    cin >> balls[i];
  }
  for (int i = 0; i < k; i++) {
    add(mat, 0, balls[i] - 1);
  }
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return;
}
int32_t main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const long long inf = 1e15;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
int large_power(int x, unsigned int y) {
  int res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
int power(int x, int y) {
  int res = 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
string bin(int n) {
  string s = "";
  while (n > 1) {
    int rem = n % 2;
    if (rem)
      s = "1" + s;
    else
      s = "0" + s;
    n = n / 2;
  }
  if (n) s = "1" + s;
  return s;
}
long long exp(long long i, long long j) {
  if (j == 1) return i;
  if (j == 0) return 1;
  if (j & 1) {
    return (i * exp((i * i) % mod, j / 2)) % mod;
  }
  return exp((i * i) % mod, j / 2) % mod;
}
long long invMod(long long i) { return exp(i, mod - 2) % mod; }
long long nCr(long long n, long long r) {
  long long ans = 1;
  for (long long i = n; i > n - r; i--) {
    ans = (ans * i) % mod;
  }
  for (long long i = 1; i <= r; i++) {
    ans = (ans * invMod(i)) % mod;
  }
  return ans;
}
tuple<int, int> divisors(int n) {
  map<int, int> mp;
  int e = 0, o = 0;
  while (n % 2 == 0) {
    if (mp[2] > 0)
      mp[2]++;
    else
      mp[2] = 1;
    n /= 2;
    e += 1;
  }
  int i = 3;
  while (i * i <= n) {
    if (n % i == 0) {
      o += 1;
      if (mp[i] > 0)
        mp[i]++;
      else
        mp[i] = 1;
      n /= i;
    } else {
      i += 2;
    }
  }
  if (n > 2) {
    o += 1;
    if (mp[n] > 0)
      mp[n]++;
    else
      mp[n] = 1;
  }
  return make_tuple(e, o);
}
int n, m, k;
vector<int> ans;
int pi = -1, pj = -1;
void add(vector<vector<int>>& mat, int row, int col) {
  if (row == n) {
    ans.push_back(col + 1);
    pi = -1;
    pj = -1;
    return;
  }
  if (pi != -1 && pj != -1) {
    mat[pi][pj] = 2;
  }
  if (mat[row][col] == 2) {
    pi = row;
    pj = col;
    add(mat, row + 1, col);
    return;
  } else if (mat[row][col] == 1) {
    pi = row;
    pj = col;
    add(mat, row, col + 1);
    return;
  } else if (mat[row][col] == 3) {
    pi = row;
    pj = col;
    add(mat, row, col - 1);
    return;
  }
}
void solve() {
  cin >> n >> m >> k;
  vector<vector<int>> mat(n, vector<int>(m));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> mat[i][j];
    }
  }
  vector<int> balls(k);
  for (int i = 0; i < k; i++) {
    cin >> balls[i];
  }
  for (int i = 0; i < k; i++) {
    add(mat, 0, balls[i] - 1);
  }
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return;
}
int32_t main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```