#include <climits>
#include <vector>
/*
lemma: I will solve subtask 2
*/
using namespace std;
double ans=0;
vector<vector<int>> visitted, exists; // both are [y][x]
vector<vector<bool>> DONE;
void bfs(int I, int x, int y, int curdist){
// coords is [x,y]
visitted[y][x] = I;
while(ans >= 1000000000){ans -= 1000000000;}
if(exists[y+1][x+1] && visitted[y+1][x+1] != I){
ans += curdist+1;
ans /= 2;
bfs(I, x+1, y+1, curdist+1);
}
if(exists[y+1][x-1] && visitted[y+1][x-1] != I){
ans += curdist+1;
ans /= 2;
bfs(I, x-1, y+1, curdist+1);
}
if(exists[y-1][x+1] && visitted[y-1][x+1] != I){
ans += curdist+1;
ans /= 2;
bfs(I, x+1, y-1, curdist+1);
}
if(exists[y-1][x-1] && visitted[y-1][x-1] != I){
ans += curdist+1;
ans /= 2;
bfs(I, x-1, y-1, curdist+1);
}
return;
}
int DistanceSum(int N, int *X, int *Y) {
DONE = vector<vector<bool>>(N+1, vector<bool>(N+1, 0));
visitted = vector<vector<int>>(N+1, vector<int>(N+1, 0));
exists = vector<vector<int>>(N+1, vector<int>(N+1, 0));
// find min X and Y values, to translate the graph to start at (1,1)
int minX=INT_MAX-1, minY=INT_MAX-1;
for(int i=0; i<N; i++){
minX = min(minX, X[i]);
minY = min(minY, Y[i]);
}
for(int i=0; i<N; i++){
exists[Y[i]-minY][X[i]-minX] = 1;
}
for(int i=0; i<N; i++){
bfs(i, X[i], Y[i], 0);
}
int realans = ans;
return realans;
}
