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>
int abs(int n)
{
	if (n < 0)
		n *= -1;
	return n;
}
int main()
{
	int n;
	scanf("%d", &n);
	int i;
	int t[100005], x[100005], y[100005];
	for (i = 1; i <= n; i++)
		scanf("%d %d %d", &t[i], &x[i], &y[i]);
	t[0] = x[0] = y[0] = 0;
	int dt, dx, dy;
	for (i = 0; i < n; i++)
	{
		dt = t[i + 1] - t[i];
		dx = abs(x[i] - x[i + 1]);
		dy = abs(y[i] - y[i + 1]);
		if (dt < dx + dy)
		{
			printf("No\n");
			return 0;
		}
		if ((dt - dx - dy) % 2 > 0)
		{
			printf("No\n");
			return 0;
		}
	}
	printf("Yes\n");
	return 0;
}