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 9.2.1) (judge verdict: Accepted)

        #include<stdio.h>
        #include <stdlib.h>
        int main(){
        	int a = 0, b = 0, N = 0, i, t[110000], x[110000], y[110000];
        	int q = 1;
        	scanf("%d", &N);
        	t[0] = 0;
        	x[0] = 0;
        	y[0] = 0;
        	for (i = 0; i < N; i++) {
        		scanf("%d %d %d", &t[i + 1], &x[i + 1], &y[i + 1]);
        	}
        	for (i = 0; i < N; i++) {
        		 a = t[i + 1] - t[i];
        		 b = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i]);
        		 if (a < b) {
        			 q = 0;
        		 }
        		 if(a % 2 != b % 2){
        			 q = 0;
        		 }
        	}
        	if (q == 1) {
        		printf("Yes");
        	}else{
              printf("No");
            }
        }