#include<bits/stdc++.h>
using namespace std;
using ll = long long;
ll query(vector<vector<ll>> &pref, int upper_row, int lower_row, int left, int right){
ll v = 0;
v += pref[upper_row][right];
if (left != 0){
v -= pref[upper_row][left - 1];
}
if (lower_row != 0){
v -= pref[lower_row - 1][right];
}
if (left != 0 && lower_row != 0){
v += pref[lower_row - 1][left - 1];
}
return v;
}
vector<ll> mosaic(vector<int> X, vector<int> Y, vector<int> T, vector<int> B, vector<int> L, vector<int> R){
vector<ll> ans;
int n = X.size();
int q = T.size();
map<pair<int, int>, bool> arr; // Vi beregner kun nogen få værdier
for (int i = 0; i < n; i++){
for (int j = 0; min(i, j) <= 3 && j < n; j++){
if (i == 0){
arr[{i, j}] = X[j];
}
else if (j == 0){
arr[{i, j}] = Y[i];
}
else{
arr[{i, j}] = 1 - (arr[{i - 1, j}] || arr[{i, j - 1}]);
}
}
}
for (int k = 0; k < n; k++){
int row = T[0];
int col = L[0];
if (row <= 3 || col <= 3){
ans.push_back(arr[{row, col}]);
}
else if (row < col){
row = 0;
col -= row;
row += 3;
col += 3;
ans.push_back(arr[{row, col}]);
}
else{
col = 0;
row -= col;
row += 3;
col += 3;
ans.push_back(arr[{row, col}]);
}
}
return ans;
}
// int main(){
// vector<ll> ans = mosaic({1, 0, 1, 0}, {1, 1, 0, 1}, {0, 2}, {0, 3}, {0, 0}, {3, 2});
// for (auto &p: ans){
// cout << p << "\n";
// }
// return 0;
// }
// Vi behøver kun de første 4 calculations
// Så hvordan fortsætter vi derfra? Jeg tror på fullsolve.
