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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;
using static System.Console;
using static System.Convert;
using static System.Math;

class Program
{
    static void Main(string[] args)
    {
        var num = ToInt32(ReadLine());
        var txys = new int[num][];
        for (var i = 0; i < num; i++)
            txys[i] = Array.ConvertAll(ReadLine().Split(' '), int.Parse);
        var isClear = true;
        if (txys[0][1] + txys[0][2] > txys[0][0]
            ||((txys[0][1]+txys[0][2])%2!=txys[0][0]%2)) { WriteLine("No");return; }
        for (var i = 0; i < num-1; i++)
        {
            var m = Abs(txys[i][1]-txys[i+1][1])+Abs(txys[i+1][2]-txys[i][2]);
            var dift = txys[i + 1][0] - txys[i][0];
            if (m > dift||m%2!=dift%2) { isClear = false;break; }
        }
        WriteLine(isClear ? "Yes" : "No");
    }
}