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: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <cmath>
#include <stack>
#include <string>
#include <iomanip>
#include <queue>
#include <deque>
#include<cstring>
#include<limits>
#include <sstream>
#include <stdio.h>
#include<unordered_map>
using namespace std;
#define sz(x) (x).size()
#define all(v) (v).begin(),(v).end()
#define m_p make_pair
void run()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#endif
}
void fast()
{
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
}
long long dist(int x, int y, int a, int b)
{
	return ((x - a)*(x - a)) + ((y - b)*(y - b));
}
int main()
{    
	fast();
	int n;
	cin >> n;
	long long t, x, y;
	long long curx = 0, cury = 0,last_t=0;
	for (int i = 0; i < n; i++)
	{
		cin >> t >> x >> y;
		long long dis = dist(curx, cury, x, y);
		if (dis <= (t-last_t)*(t-last_t)&&(curx!=x||cury!=y))
		{
			curx = x; cury = y;
			last_t = t;
		}
		else
			return cout << "No" << endl,0;
	}
	cout << "Yes" << endl;
	return 0;

}