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>
#include <string.h>
#include <math.h>
 

 
int main()
{
    int N;
    int  t[100002] ={};
    int x[100002] ={};
    int y[100002] ={};
   int kai;
   int i;
    int t_temp=0;
    int x_temp=0;
    int y_temp=0;
  
    t[0]=0;
    x[0]=0;
    y[0]=0;
  
 // 整数の入力
    scanf("%d", &N);
 // printf("%d\n",N);
    for(i = 1; i < N +1;i++){
      scanf("%d %d %d", &t[i],&x[i], &y[i]);      
//      printf("%d %d %d", t[i],x[i], y[i]);
      t_temp=t[i] - t[i-1];
/*      if(0 < x[i] - x[i-1]) x_temp=x[i] - x[i-1];
      else  x_temp=x[i-1] - x[i];
      if(0 < y[i] - y[i-1]) y_temp=y[i] - y[i-1];
      else  y_temp=y[i-1] - y[i];
 */
      x_temp=abs(x[i] - x[i-1]);
      y_temp=abs(y[i] - y[i-1]);
      
//      printf("%d %d %d\n", t_temp,x_temp, y_temp);
 //     printf("%d \n", abs((t_temp-(x_temp+y_temp))));
  //    printf("%d \n", t_temp-(x_temp+y_temp));
      if((0>(t_temp-(x_temp+y_temp))) || ((1==(t_temp-(x_temp+y_temp)) %2))){
//      if(0>(t_temp-(x_temp+y_temp))){
//        if(0 != abs((t_temp-(x_temp+y_temp)) ) %2 ){
        printf("No\n");
        return 0;
      }
    }
 
    printf("Yes\n");

return 0;
}