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 (Clang 3.8.0) (judge verdict: Accepted)

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

int main()
{
    int N = 0;
    scanf("%d",&N);
    struct point{
        int timet;
        int xt;
        int yt;
    };
    struct point *information = (struct point *)malloc((N + 1) * sizeof(struct point));
    information[0].timet = 0;
    information[0].xt = 0;
    information[0].yt = 0;
    for(int i = 1;i <= N;i++)
    {
        scanf("%d",&(information[i].timet));
        scanf("%d",&(information[i].xt));
        scanf("%d",&(information[i].yt));
    }
    int can_or_not = 1;
    for(int i = 1;i <= N && can_or_not == 1;i++)
    {
        if(information[i].timet - information[i - 1].timet < abs(information[i].xt - information[i - 1].xt) + abs(information[i].yt - information[i - 1].yt))
        {
            can_or_not = 0;
            break;
        }
        else
        {
            if( (information[i].timet - information[i - 1].timet)  % 2 != (abs(information[i].xt - information[i - 1].xt) + abs(information[i].yt - information[i - 1].yt)) % 2)
            {
                can_or_not = 0;
                break;
            }
            
        }
    }
    if(can_or_not == 1)
    {
        printf("Yes");
    }
    else if(can_or_not == 0)
    {
        printf("No");
    }
    free(information);
    return 0;
}