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>
#include<stdlib.h>

int main(void){
    int N;
    scanf("%d",&N);
    int i,t[100001], x[100001], y[100001];
    int dt,dx,dy,d;

    t[0] = 0;
    x[0] = 0;
    y[0] = 0;

    for(i = 1;i < N+1;i++) scanf("%d%d%d",&t[i],&x[i],&y[i]);
    
    for(i = 1;i < N+1;i++){
        dt = t[i] - t[i-1];
        dx = x[i] - x[i-1];
        dy = y[i] - y[i-1];

        d = abs(dx) + abs(dy);
        
        if(d > dt){
            printf("No");
            return 0;
        }else{
            if((dt - d) % 2 != 0){
                printf("No");
                return 0;
            }
        }
    }
    
    printf("YES");

    return 0;
}