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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

void solve(void)
{
	int n;
	int x0, y0, t0;
	int x1, y1, t1;
	int i;
	char ngTxt[] = "No", okTxt[] = "Yes", *pTxt = okTxt;
	
	x1 = y1 = t1 = 0;
	
	scanf("%d", &n);
	for(i=0;i<n;++i){
		int d;
		x0 = x1;
		y0 = y1;
		t0 = t1;
		
		scanf("%d %d %d", &t1, &x1, &y1);
		d = (x1>x0 ? x1-x0 : x0-x1) + (y1>y0 ? y1-y0 : y0-y1);
		if( d > t1 - t0 ){
			pTxt = ngTxt;
			break;
		}
		if( (t1-t0-d)%2 == 1 ){
			pTxt = ngTxt;
			break;
		}
	}
	printf("%s\n", pTxt);
}
int main(int argc, char **argv)
{
	solve();
	return 0;
}