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>

int diff(int a, int b){
	return a - b > 0 ? a - b : b - a;
}

int main(void){
	int n;
	scanf("%d", &n);
	int t[n], x[n], y[n];
	for(int i = 0; i < n; i++){
		scanf("%d %d %d", &t[i], &x[i], &y[i]);
	}

	if((t[0] < x[0] + y[0]) || diff(t[0], x[0] + y[0]) % 2 != 0){
		printf("No\n");
		return 0;
	}

	for(int i = 0; i < n - 1; i++){
		int timed = t[i + 1] - t[i];
		int moved = diff(x[i + 1], x[i]) + diff(y[i + 1], y[i]);
		if(timed == moved){
			continue;
		}else if(timed >= diff(timed, moved) && diff(timed, moved) % 2 == 0){
			continue;
		}else{
			printf("No\n");
			return 0;
		}
	}
	printf("Yes\n");
	return 0;
}