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

int manhattan(int x1,int y1,int x2,int y2);

int main()
{
	int n,i,m,tsa,flag=1;
	int t[100010][3]={0};
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%d %d %d",&(t[i][0]),&(t[i][1]),&(t[i][2]));
		m=manhattan(t[i-1][1],t[i-1][2],t[i][1],t[i][2]);
		tsa=t[i][0]-t[i-1][0];
		if(tsa<m||(tsa-m)%2==1)flag=0;
	}
	if(flag)printf("Yes");
	else printf("No");
}

int manhattan(int x1,int y1,int x2,int y2)
{
	return (abs(x1-x2)+abs(y1-y2));
}