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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

int main(){
	
	int N,t[100010],c[100010][3],i,z;
	t[0]=0;
	c[0][1]=0;
	c[0][2]=0;
	
	scanf("%d",&N);
	
	for(i=1;i<=N;i++){
		scanf("%d",&t[i]);
		scanf("%d",&c[i][1]);
		scanf("%d",&c[i][2]);
	}
	
	for(i=1;i<=N;i++){
		z=abs(c[i][1]-c[i-1][1])+abs(c[i][2]-c[i-1][2]);
		if((t[i]-t[i-1]-z)<0||(t[i]-t[i-1]-z)%2==1){
			printf("\nNo\n\n");
			return 0;
		}
	}
	
	printf("\nYes\n\n");
	
	
	return 0;
}