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 main(void)
{
  int num,count=0;
  scanf("%d",&num);
  int i, t[num], x[num], y[num],dt,dx,dy;
  for(i=0; i<num; i++)
  {
    scanf("%d%d%d", &t[i],&x[i],&y[i]);
  }
  dt=t[0];
  dx=x[0];
  dy=y[0];
  if(dx<0) dx=-dx;
  if(dy<0) dy=-dy;
  if(dx+dy>dt || (dt-(dx+dy))&2==1) count=1;
  for(i=1; i<num; i++)
  {
    dt=t[i]-t[i-1];
    dx=x[i]-x[i-1];
    dy=y[i]-y[i-1];
    if(dx<0) dx=-dx;
    if(dy<0) dy=-dy;
    if(dx+dy>dt || (dt-(dx+dy))&2==1) count=1;
  }
  if(count==0) printf("Yes");
  else printf("No");
  return 0;
}