Vasya has the square chessboard of size n × n and m rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.

The cell of the field is under rook's attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.

You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ min(100 000, n²)) — the size of the board and the number of rooks.

Each of the next m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n) — the number of the row and the number of the column where Vasya will put the i-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.

Print m integer, the i-th of them should be equal to the number of cells that are not under attack after first i rooks are put.

On the picture below show the state of the board after put each of the three rooks. The cells which painted with grey color is not under the attack.

题意：n*n的棋盘  m个rook（车 象棋中的ju）  给你m的车的坐标 车所在的行列不能摆放棋子

每摆放一个车 输出还能摆放其他旗子的位置的数量  具体看样例

题解：每次增加一个车 都更新不能摆放棋子的行列的个数 ro，cl

注意行列的交叉点会重复计算 记录每次的结果为ans=n*n-(ro+cl)*n+ro*cl;

 //code  by drizzle

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #define ll __int64

 #define PI acos(-1.0)

 #define mod 1000000007

 using namespace std;

 ll n,m;

 ll h[];

 ll l[];

 ll cl,ro;

 ll x,y;

 ll ans[];

 int main()

 {

     cl=;

     ro=;

     scanf("%I64d %I64d",&n,&m);

     memset(l,,sizeof(l));

     memset(h,,sizeof(h));

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

     {

         scanf("%I64d %I64d",&x,&y);

         if(h[x]==)

         {

             h[x]=;

             ro++;

         }

         if(l[y]==)

         {

             l[y]=;

             cl++;

         }

         ans[i]=n*n-(ro+cl)*n+ro*cl;

     }

     printf("%I64d",ans[]);

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

         printf(" %I64d",ans[i]);

     return ;

 }