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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int N,yn=1,i;
    int *t = (int *)malloc(sizeof(int)*N);
    int *x = (int *)malloc(sizeof(int)*N);
    int *y = (int *)malloc(sizeof(int)*N);

    scanf("%d",&N);
    for(i=0;i<N;i++){
        scanf("%d %d %d",&t[i],&x[i],&y[i]);
    }

    if(abs(t[0]-x[0]-y[0])%2 == 1 || t[0]<x[0]+y[0]) yn = 0;

    for(i=0;i<N-1;i++){
        if(abs((t[i+1]-t[i])-abs(x[i+1]-x[i])-abs(y[i+1]-y[i]))%2 == 1 || (t[i+1]-t[i])<((x[i+1]-x[i])+(y[i+1]-y[i]))){
            yn = 0;
            break;
        }
    }

    if(yn == 1) printf("Yes\n");
    else printf("No\n");

    return 0;
}