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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int myabs(int n)
{
	return n>0?n:-n;
}

int main()
{
	int n, i;
	scanf("%d", &n);
	int flag = 1;
	int t0 = 0, x0 = 0, y0 = 0;
	for(i=0;i<n;i++)
	{
		int t, x, y;
		scanf("%d%d%d", &t, &x, &y);
		int dt = myabs(t-t0);
		int dx = myabs(x-x0);
		int dy = myabs(y-y0);
		if((dx+dy)%2!=dt%2)
			flag = 0;
		else if((dx+dy)>dt)
			flag = 0;
		t0 = t;
		x0 = x;
		y0 = y;
	} 
	if(flag)
		printf("Yes\n");
	else
		printf("No\n");
    return 0;
}