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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

#define TRUE 1
#define FALSE 0

int reachable(int t_b, int x_b, int y_b,
	int t_a, int x_a, int y_a) {
	int dt = t_a - t_b, dx = abs(x_a - x_b), dy = abs(y_a - y_b);
	if (dt % 2 != (dx % 2 + dy % 2) % 2 ) {
		return FALSE;
	}
	if (dx + dy > dt) {
		return FALSE;
	}
	return TRUE;
}

int main(void) {
	int x_before = 0, y_before = 0, t_before = 0, n, sw,
		x_after, y_after, t_after;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d %d %d", &t_after, &x_after, &y_after);
		sw = reachable(t_before, x_before, y_before,
			t_after, x_after, y_after);
		if (sw) {
			t_before = t_after;
			x_before = x_after;
			y_before = y_after;
		}
		else if (!sw) {
			break;
		}
	}
	if (sw) {
		printf("YES\n");
	}
	else if (!sw) {
		printf("NO\n");
	}
	return 0;
}