Codeforces Round #614 C.NEKO's Maze Game

Codeforces Round #614 C.NEKO’s Maze Game

NEKO#ΦωΦ has just got a new maze game on her PC!

The game’s main puzzle is a maze, in the forms of a 2×n rectangle grid. NEKO’s task is to lead a Nekomimi girl from cell (1,1) to the gate at (2,n) and escape the maze. The girl can only move between cells sharing a common side.

However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.

After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (ri,ci) (either from ground to lava or vice versa).

Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1,1) to cell (2,n) without going through any lava cells.

Although NEKO is a great streamer and gamer, she still can’t get through quizzes and problems requiring large amount of Brain Power. Can you help her?

Input

The first line contains integers n, q $(2≤n≤10^5, 1≤q≤10^5)$.

The i-th of q following lines contains two integers ri, ci (1≤ri≤2, 1≤ci≤n), denoting the coordinates of the cell to be flipped at the i-th moment.

It is guaranteed that cells (1,1) and (2,n) never appear in the query list.

Output

For each moment, if it is possible to travel from cell (1,1) to cell (2,n), print “Yes”, otherwise print “No”. There should be exactly q answers, one after every update.

You can print the words in any case (either lowercase, uppercase or mixed).

Example

input

5 5
2 3
1 4
2 4
2 3
1 4

output

Yes
No
No
No
Yes

个人认为不难，对 $(1,a)$和 $(2,b)$，只要 $|a-b| \le1$，那么 $(a,b)$即可构成一组障碍，那么只需实施统计当其障碍组数即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int a[2][100005],cnt=0,n,q;
void judge(int r,int c){
    for(int i=-1;i<=1;i++){
        if(i+c<1 && i+c>n) continue;
        if(a[r^1][i+c]==1 && a[r][c]) cnt++;
        else if(a[r^1][i+c]==1 && !a[r][c]) cnt--;
    }
}
int main()
{
    memset(a,0,sizeof(a));
    cin>>n>>q;
    int r,c;
    while(q--){
        cin>>r>>c;
        a[r-1][c]^=1;
        judge(r-1,c);
        if(!cnt) cout<<"yes"<<endl;
        else cout<<"no"<<endl;
    }
    return 0;
}