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>

int main(void){
  int n, place_sa, x, y;
  int yes_flag = 1;

  scanf("%d", &n);
  
  int zahyou[3*(n + 1)];
  
  for(int i = 0; i < 3; i++){
    zahyou[i] = 0;
  }

  for(int i = 1; i <= n; i++){
    scanf("%d %d %d", &zahyou[3*i+0], &zahyou[3*i+1], &zahyou[3*i+2]);
  }
  for(int i = n; i > 0; i--){
    x = zahyou[3*i+1] - zahyou[3*(i - 1)+1];
    place_sa = zahyou[3*i+0] - zahyou[3*(i - 1)+0];
    if(x < 0) {
        x = -x;
        }
    y = zahyou[3*i+2] - zahyou[3*(i - 1)+2];

    if(y < 0){ 
        y = -y;
        }
    if((place_sa % 2) != ((x + y) % 2) || (x + y) > place_sa){
      yes_flag = 0;
      break;
    }
  }
  if(yes_flag == 1){
    printf("Yes\n");
  }else{
    printf("No\n");
  }
  return 0;
}