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: Runtime Error)

#include <stdio.h>


int main()
{
	int n, t[10000], x[10000], y[10000];
	int j, k = 0;
	scanf("%d",&n);
	for (j = 0; j < n; j++) {
		scanf("%d %d %d", &t[j], &x[j], &y[j]);
		if (j > 0) {
			if (((x[j - 1] + 1) == x[j] && y[j-1] == y[j]) || ((x[j - 1] - 1) == x[j] && y[j - 1] == y[j]) || ((y[j - 1] + 1) == y[j] && x[j - 1] == x[j]) || ((y[j - 1] - 1) == y[j] && x[j - 1] == x[j]))
				k = 1;
			else
				k = 0;
		}
	}
	if (k == 1)
		printf("Yes\n");
	else
		printf("No\n");

    return 0;
}