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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include<stdio.h>
#include<math.h>
int main()
{
    int m,n,s,v,i,j,a[10002][2],p,q,b[10002];
    while(~scanf("%d",&n))
    {
        v=1;
        for(i=0; i<n; i++)
            scanf("%d %d %d",&b[i],&a[i][0],&a[i][1]);
        p=0;
        q=0;
        for(i=0; i<n; i++)
        {
            s=abs(a[i][0]-p)+abs(a[i][1]-q);
            if(i>0)
                b[i]=b[i]-b[i-1];
            if(s>b[i]||(b[i]-s)%2!=0)
            {
                v=0;
                break;
            }
            p=a[i][0];
            q=a[i][1];
        }
        if(v==1)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}