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>
#include <stdlib.h>
#include <math.h>

typedef struct {
	int time;
	int x;
	int y;
} coord;

int time_sort(const void* pt1, const void* pt2) {
	return ((coord*)pt1)->time - ((coord*)pt2)->time;
}

int main() {
	int N, i, d, t, flg = 0;
	coord pts[100010];
	scanf("%d", &N);
	for (i = 0; i < N; i++) scanf("%d%d%d", &pts[i].time, &pts[i].x, &pts[i].y);
	pts[N].time = pts[N].x = pts[N].y = 0;
	qsort(pts, N + 1, sizeof(coord), time_sort);
	for (i = 1; i <= N; i++) {
		d = abs(pts[i].x - pts[i - 1].x) + abs(pts[i].y - pts[i - 1].y);
		t = pts[i].time - pts[i - 1].time;
		if (d > t || (t - d) % 2 != 0) {
			flg = 1;
			break;
		}
	}
	printf("%s", flg ? "No" : "Yes");
	return 0;
}