Problem:
Score : 300 points

Problem Statement

AtCoDeer the deer is going on a trip in a two-dimensional plane.
In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.

If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1).
Note that he cannot stay at his place.
Determine whether he can carry out his plan.

Constraints

1 ≤ N ≤ 10^5

0 ≤ x_i ≤ 10^5

0 ≤ y_i ≤ 10^5

1 ≤ t_i ≤ 10^5

t_i < t_{i+1} (1 ≤ i ≤ N-1)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t_1 x_1 y_1
t_2 x_2 y_2
:
t_N x_N y_N

Output

If AtCoDeer can carry out his plan, print Yes; if he cannot, print No.

Sample Input 1

2
3 1 2
6 1 1

Sample Output 1

Yes

For example, he can travel as follows: (0,0), (0,1), (1,1), (1,2), (1,1), (1,0), then (1,1).

Sample Input 2

1
2 100 100

Sample Output 2

No

It is impossible to be at (100,100) two seconds after being at (0,0).

Sample Input 3

2
5 1 1
100 1 1

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cmath>

using namespace std;


int main(){

	int N,x[2],y[2],time[2];

	cin >> N;

	cin >> time[0]>>x[0]>>y[0]; 

	bool can=0;
	for(int i=1; i<N; i++){

		cin >> time[1]>>x[1]>>y[1]; 

		if(abs( x[1]-x[0] )+ abs(y[1]-y[0] ) <= time[1]-time[0]  && (abs( x[1]-x[0] )+abs(y[1]-y[0] ))%2 == (time[1]-time[0])%2){

			can=1;

		}
		//cout<<abs(x[1]-x[0] )+abs(y[1]-y[0])<<" "<<time[1]-time[0] <<" "<<(abs(x[1]-x[0]) +abs(y[1]-y[0]) )%2<<" "<< (time[1]-time[0])%2<<endl;

		time[0]=time[1];
		x[0]=x[1];
		y[0]=y[1];

	}

	if(can) cout <<"Yes" <<endl;
		
	else cout<<"No"<<endl;





}