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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

int main() {
	int N, i, x_dif, y_dif, t_dif, flag;
	int x[100000];
	int y[100000];
	int t[100000];

	x[0] = 0;
	y[0] = 0;
	t[0] = 0;
	scanf("%d", &N);

	for (i = 0; i < N; i++) {
		scanf("%d %d %d", &t[i+1], &x[i+1], &y[i+1]);
	}

	for (i = 0; i < N; i++) {
		x_dif = abs(x[i + 1] - x[i]);
		y_dif = abs(y[i + 1] - y[i]);
		t_dif = abs(t[i + 1] - t[i]);
		if (x_dif + y_dif <= t_dif && abs((x_dif-y_dif)%2)==t_dif%2)flag = 0;
		else flag = 1;
		if (flag == 1)break;
	}
	if (flag == 0)printf("Yes\n");
	else printf("No\n");
	return 0;
}