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 <bits/stdc++.h>
#define N 1000005
#define ll long long int 	
#define MP make_pair
#define pb push_back
#define ppb pop_back
#define sp " "
#define endl "\n"
#define fi first
#define se second
#define ii pair<int,int>
#define lli pair<ll,ll>
#define fast cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(false)
#define fast2 freopen ("myfile.in","r",stdin);freopen ("myfile.out","w",stdout);
#define mod 1000000007
#define fs(x,y) for(int i=1;i<=y;i++) cin>>x[i];
#define fo(i,x,y) for(int i=x;i<=y;i++)
using namespace std;

ll n,m,ar[N],sum;


int main()
{
	fast;
	cin>>n;
	int a=0;
	int b=0;
	int q=0;
	int flag=0;
	for(int i=1;i<=n;i++)
	{
		int t,l,r;
		cin>>t>>l>>r;
		int x=fabs(l-a);
		int y=fabs(r-b);
		if(x+y>t-q || (t-q-x-y)%2==1)
			flag=1;
		a=l;
		b=r;
		q=t;
	}
	if(flag)
	{
		return cout<<"No",0;
	}
	cout<<"Yes";

}