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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php

fscanf(STDIN, '%d', $n);
$input = [];
for ($i = 0;$i < $n; $i++){
    $input[] = explode(' ',rtrim(fgets(STDIN), PHP_EOL));
}

$before_t = 0;
$before_x = 0;
$before_y = 0;


foreach ($input as $value) {
    $diff_t = abs($value[0] - $before_t);
    $diff_p = abs(($value[1]-$before_x) + ($value[2] - $before_y));

//    var_dump($diff_t, $diff_p,$diff_t < $diff_p);
    if ($diff_t < $diff_p) {
        echo 'No' . PHP_EOL;
        return;
    }

    if ($diff_t % 2 !== 0 && $diff_p %2 === 0) {
        echo 'No'.PHP_EOL;
        return;
    } elseif ($diff_t %2 === 0 && $diff_p %2 !== 0) {
        echo 'No'.PHP_EOL;
        return;
    }

    $before_t = $value[0];
    $before_x = $value[1];
    $before_y = $value[2];
}
echo 'Yes'.PHP_EOL;