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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string>
#include <stdlib.h>
#include <queue>
#include <stack>
#include <utility>
#include <fstream>
#include <random>
#include <map>
#include <unordered_map>
#include <cstdlib>
#define rep(i,n) for(int i=0;i<n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define ll long long
#define INF 1000000001
#define mod 1000000007
#define p pair<int,int>
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
using namespace std;
int POW(int x,int y){return int(pow(double(x),double(y)));}


int main(){
	int n;
	cin>>n;
	vector<int> t(n),x(n),y(n);
	rep(i,n){
		cin>>t[i]>>x[i]>>y[i];
	}
	if(t[0]<x[0]+y[0] || (t[0]-(x[0]+y[0]))%2==1){
		cout<<"No"<<endl;
		return 0;
	}
	rep(i,n-1){
		if(t[i+1]-t[i] < abs(x[i+1]-x[i])+abs(y[i+1]-y[i]) || ((t[i+1]-t[i])-(abs(x[i+1]-x[i])+abs(y[i+1]-y[i])))%2==1){
			cout<<"No"<<endl;
			return 0;
		}
	}
	cout<<"Yes"<<endl;
	return 0;
}