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>
#include <stdlib.h>

int cango(int t1, int t2, int x1, int x2, int y1, int y2){
    if((int)(abs(x2-x1)+abs(y2-y1))%2==0){
        if(t2-t1>=(int)(abs(x2-x1)+abs(y2-y1))&&(t2-t1)%2==0) return 1;
        else return 0;
    } else {
        if(t2-t1>=(int)(abs(x2-x1)+abs(y2-y1))&&(t2-t1)%2==1) return 1;
        else return 0;
    }
}

int main()
{
    int N;
    scanf("%d", &N);
    int t[N+1], x[N+1], y[N+1];
    int i;
    for(i=1;i<=N;i++) scanf("%d %d %d", &t[i], &x[i], &y[i]);
    
    int judge = 1;
    judge *= cango(0, t[1], 0, x[1], 0, y[1]);
    for(i=1;i<N;i++){
        judge *= cango(t[i], t[i+1], x[i], x[i+1], y[i], y[i+1]);
    }
    
    if(judge==1) printf("Yes");
    else printf("No");

    return 0;
}