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>
#define abs(a) ((a)<0 ? -(a) : (a))
int main(){
  int n;  scanf("%d", &n);

  int nowt = 0, nowx = 0, nowy = 0;
  int nt, nx, ny;
  for(int i=0; i<n; i++){
    scanf("%d%d%d", &nt, &nx, &ny);
    if(nx == nowx && ny == nowy){
      if((nt-nowt)%2){
	printf("No\n");
	return 0;
      }
    }
    if(!(nt - nowt == abs(nx-nowx)+abs(ny-nowy) || (nt - nowt > abs(nx-nowx)+abs(ny-nowy) && (nt - nowt)%2==0))){
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  return 0;
}