#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < (n); i++)
#define repr(i, n) for (int i = (n) - 1; i >= 0; i--)

using namespace std;
using ll = long long;

int dp[50][250][250];
ll ep[250][250];

constexpr ll inf = 1e13;

void chmin(ll &x, ll y) {
  x = min(x, y);
}

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int H, W, GY, GX; cin >> H >> W >> GY >> GX;
  GY--; GX--;
  vector<vector<int>> A(H, vector<int>(W));
  rep(i, H) rep(j, W) cin >> A[i][j];
  rep(k, 50) rep(i, 250) rep(j, 250) dp[k][i][j] = 1e9;
  dp[0][GY][GX] = A[GY][GX];
  const int dy[] = {0, 1, 0, -1};
  const int dx[] = {1, 0, -1, 0};
  rep(k, 50) {
    dp[k][GY][GX] = A[GY][GX] + k*k;
    priority_queue<tuple<ll, int, int>> q;
    q.emplace(-dp[k][GY][GX], GY, GX);
    while (!q.empty()) {
      ll d;
      int y, x;
      tie(d, y, x) = q.top(); q.pop();
      if (-d > dp[k][y][x]) continue;
      rep(l, 4) {
        int ny = y + dy[l];
        int nx = x + dx[l];
        if (0 <= ny && ny < H && 0 <= nx && nx < W) {
          if (dp[k][ny][nx] > dp[k][y][x] + A[ny][nx] + k*k) {
            dp[k][ny][nx] = dp[k][y][x] + A[ny][nx] + k*k;
            q.emplace(-dp[k][ny][nx], ny, nx);
          }
        }
      }
    }
  }
  ep[GY][GX] = A[GY][GX] + inf;
  rep(i, 250) rep(j, 250) ep[i][j] = 1e18;
  priority_queue<tuple<ll, int, int>> q;
  q.emplace(-ep[GY][GX], GY, GX);
  while (!q.empty()) {
    ll d;
    int y, x;
    tie(d, y, x) = q.top(); q.pop();
    if (-d > ep[y][x]) continue;
    rep(l, 4) {
      int ny = y + dy[l];
      int nx = x + dx[l];
      if (0 <= ny && ny < H && 0 <= nx && nx < W) {
        if (ep[ny][nx] > ep[y][x] + A[ny][nx] + inf) {
          ep[ny][nx] = ep[y][x] + A[ny][nx] + inf;
          q.emplace(-ep[ny][nx], ny, nx);
        }
      }
    }
  }
  int Q; cin >> Q;
  while (Q--) {
    int y, x;
    ll k; cin >> y >> x >> k; y--; x--;
    if (k < 50) {
      cout << dp[k][y][x] << '\n';
    } else {
      ll d = (abs(GY - y) + abs(GX - x) + 1);
      cout << ep[y][x] - d * inf + d * k * k << '\n';
    }
  }
}