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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

// ABC 086

// inputに入力データ全体が入る
function Main(input) {
  var arr = input.split("\n");
  var x = 0;
  var y = 0;
  var result = "No";
  for(var i=1;i<arr.length;i++) {
    var tmp = arr[i].split(" ");
    var x_range = check(x,tmp[1]); //現在地から行く地点までの差 x
    var y_range = check(y,tmp[2]); //現在地から行く地点までの差 y

      // 時間は余っているが、到着地点に達した場合
      if(x_range+y_range < tmp[0]) {

        // 残り時間を持て余せない場合はNo
        if(tmp[0]-(x_range+y_range) % 2 == 0)result = "No";

        // 時間が足りない場合
      } else if(x_range+y_range > tmp[0]) {
        result = "No";
      }

    x = tmp[1];
    y = tmp[2];
  }
  console.log(result);
}

function check(before,after) {

  // プラス方向に進んだ場合
  if(before < after) {
    return after - before;
  } else {
    return before - after;
  }
};

Main(require("fs").readFileSync("/dev/stdin", "utf8"));