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>

int judge(int *);

int main(void){
    
    int n, p[3] = {0}, p1[3] = {0} , flg = 0;
    scanf("%d", &n);

    scanf("%d %d %d", &p[0], &p[1], &p[2]);
    if(judge(p) == 0){
            flg = 1;
    }
    
    while(n - 1){
      scanf("%d %d %d", &p1[0], &p1[1], &p1[2]);
      p[0] = p1[0];
      p[1] = abs(p[1] - p1[1]);
      p[2] = abs(p[2] - p1[2]);
      
      if(judge(p) == 0){
	flg = 1;
      }
      p[0] = p1[0];
      p[1] = p1[1];
      p[2] = p1[2];
      
        n--;
    }

    if(flg == 0)
        printf("Yes\n");
    else
        printf("No\n");


    return 0;
}

int judge(int *p){

    int sum = 0;
    double a, b;

    a = p[1] * p[1];
    b = p[2] * p[2];

    sum = p[1] + p[2];

    if((p[0] >= sum) && !( p[0] % sum))
        return 1;
    else
        return 0;
}