#include<stdio.h>

 #include<queue>

 #include<iostream>

 using namespace std;

 typedef pair<int, int> P;

 const int maxn =  + ;    //点的最大个数

 const int SIZE =  + ;        //地图大小

 const int offset = ;        //偏置

 const int dx[] = { -,,, };

 const int dy[] = { ,,,- };

 int n,xi[maxn],yi[maxn],ans[maxn];

 bool vis[SIZE][SIZE], maze[SIZE][SIZE];

 int bfs(int x, int y)

 {

     int ret = ;

     queue<P>q;

     vis[x][y] = true;

     q.push(P(x, y));

     while (!q.empty())

     {

         P p = q.front(); q.pop();

         ret++;

         for (int i = ; i < ; i++)

         {

             int nx = p.first + dx[i], ny = p.second + dy[i];

             if (nx >=  && nx < SIZE && ny >=  && ny < SIZE && (!vis[nx][ny]))

             {

                 vis[nx][ny] = true;

                 q.push(P(nx, ny));

             }

         }

     }

     return ret;

 }

 int main()

 {

     scanf("%d", &n);

     for(int i = ;i <= n;i++)

     {

         scanf("%d%d", &xi[i], &yi[i]);

         maze[xi[i] + offset][yi[i] + offset] = ;

         vis[xi[i] + offset][yi[i] + offset] = ;

     }

     bfs(, );

     int res = ;

     for (int i = ; i < SIZE; i++)

         for (int j = ; j < SIZE; j++)

             if (vis[i][j] == false)  res++;

     ans[n] = res;

     for (int i = n; i >= ; i--)  //1~3棵以内不可能围成空地

     {

         int u = xi[i] + offset;

         int v = yi[i] + offset;

         maze[u][v] = ;

         int cnt = ;

         for (int j = ; j < ; j++)

         {

             int nu = u + dx[j], nv = v + dy[j];

             if (maze[nu][nv] ==  && vis[nu][nv] == )  cnt++;

         }

         if (cnt == )   //cnt == 0,表示该点在内部

         {

             vis[u][v] = ;

             res++;

         }

         else res -= (bfs(u, v) - );

         ans[i - ] = res;

     }

     for (int i = ; i <= n; i++)

         printf("%d\n", ans[i]);

     return ;

 }