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>


int traveling(int t, int startx, int starty, int *dest, int *move);

int move_0[2] = {1, 0};
int move_1[2] = {-1, 0};
int move_2[2] = {0, 1};
int move_3[2] = {0, -1};

int main(int argc, char const* argv[])
{
	int n, ct, pt = 0, x, y;
	int start[2] = {0, 0};
	int dest[2];

	scanf("%d", &n);

	for(int i = 0; i < n; i++)
	{
		scanf("%d %d %d", &ct, &x, &y);
		dest[1] = x;
		dest[0] = y;
		if(!traveling(ct - pt, start[1], start[0], dest, NULL))
		{
			puts("No");
			return 0;
		}
		start[1] = dest[1];
		start[0] = dest[0];
		pt = ct;
	}
	puts("Yes");
	return 0;
}

int traveling(int t, int startx, int starty, int *dest, int *move)
{
	int nextx, nexty;
	if(t < 0)
		return 0;

	if(move != NULL)
	{
		int d[2];

		d[1] = dest[1] - (startx + move[1]);
		d[0] = dest[0] - (starty + move[0]);
		
		if(d[1] < 0 || d[0] < 0 || d[1] > 100000 || d[0] > 100000)
			return 0;
		if((d[1] == 0) && (d[0] == 0) && (t == 0))
			return 1;


		nextx = d[1];
		nexty = d[0];
	} else {
		nextx = startx;
		nexty = starty;
	}

	if(traveling(t - 1, nextx, nexty, dest, move_0))
		return 1;
	if(traveling(t - 1, nextx, nexty, dest, move_1))
		return 1;
	if(traveling(t - 1, nextx, nexty, dest, move_2))
		return 1;
	if(traveling(t - 1, nextx, nexty, dest, move_3))
		return 1;
}