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>

int check(int a,int t[100000],int x[100000],int y[100000])
{
    int time,ntfx,ntfy;
    if(a>0)
    {
        time = t[a] - t[a-1];
        ntfx = x[a] - x[a-1];
        ntfy = y[a] - y[a-1];
    }
    else
    {
        time = t[a];
        ntfx = x[a];
        ntfy = y[a];
    }
    int judge = time - (ntfx+ntfy);
    if(judge<0||judge%2==1)
    {
        return 0;
    }
    return 1;
}

int main(void)
{
    int n,t[100000],x[100000],y[100000];
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d%d",&t[i],&x[i],&y[i]);
    }
    for(int i=0;i<n;i++)
    {
        if(check(i,t,x,y)==0)
        {
            printf("No");
            return 0;
        }
    }
    printf("Yes");
    return 0;
}