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>
#include <math.h>

#define MAX 100010

typedef struct point{
  int t;
  int x;
  int y;
} point;

int isMovable(const point *, const point *);

int main(){
  int n;
  point p[MAX+1] = {};
  p[0].t = 0;
  p[0].x = 0;
  p[0].y = 0;
  int i;
  scanf("%d", &n);
  if(n < 1 || MAX < n)
    return 1;
  for(i = 0; i < n; i++){
    scanf("%d %d %d", &p[i+1].t, &p[i+1].x, &p[i+1].y);
    if( p[i+1].t < 1 || MAX < p[i+1].t || p[i+1].x < 1 || MAX < p[i+1].x || p[i+1].y < 1 || MAX < p[
i+1].y )
      return 1;
    if (i > 0 && p[i+1].t < p[i].t)
      return 1;
  }
  
  for (i = 1; i < n+1; i++){
    if(isMovable(&p[i], &p[i-1])){
      printf("No\n");
      return 1;
    }
  }

  printf("Yes\n");
  return 0;
}

int
isMovable(const point *after, const point *before){
  int duration = (after -> t) - (before -> t);
  int dist = abs(after -> x - before -> x) + abs(after -> y - before -> y);
  if(dist > duration)
    return 1;
  if( dist %2 == duration %2 )
    return 0;
  else
    return 1;
}