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)

#pragma warning(disable:4996)
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define MAX_STRING 30

	int anTime[100001];
	int anX[100001];
	int anY[100001];

int main() {
	char sInput[MAX_STRING+1]="";
	char* psInput;

	int i=0;
	int nPointNumber=0;
	int nFailed=0;
	int nTimeCounter=1;
	int nLeftTime=0;
	int nXDistance=0;
	int nYDistance=0;
	int nSumDistance=0;
	int nTempNumber=0;

	// read line 1
	gets(sInput);
	nPointNumber = atoi(sInput);

	anTime[0]=0;
	anX[0]=0;
	anY[0]=0;

	// read line 2 ~
	for (i=1;i<=nPointNumber;i++){
		gets(sInput);
		psInput = strtok(sInput," ");
		anTime[i] = atoi(psInput);
		psInput = strtok(NULL," ");
		anX[i] = atoi(psInput);
		psInput = strtok(NULL," ");
		anY[i] = atoi(psInput);
	}

	while(nFailed==0 && nTimeCounter<=nPointNumber){
		nLeftTime=anTime[nTimeCounter]-anTime[nTimeCounter-1];
		nXDistance=abs(anX[nTimeCounter]-anX[nTimeCounter-1]);
		nYDistance=abs(anY[nTimeCounter]-anY[nTimeCounter-1]);
		nSumDistance=nXDistance+nYDistance;

		if ( nLeftTime < nSumDistance ){
			nFailed=1;
		} else {
			nTempNumber = nLeftTime - nSumDistance;
			if ( nTempNumber % 2 == 1 ){
				nFailed=1;
			} 
		}

		nTimeCounter++;
	}


	if (nFailed>0){
		printf("No\n");
	} else {
		printf("Yes\n");
	}

    return 0;
}