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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {
        n: usize,
        n1:[(i32,(i32,i32));n],
    }
    let mut tmp = (0, (0, 0));
    for x in 0..n {
        let tmp1: (i32, i32) = n1[x].1;
        let tmpx = tmp.1;
        if (tmp1.0 - tmpx.0).abs() + (tmp1.1 - tmpx.1).abs() <= (n1[x].0 - tmp.0) {
            if ((tmp1.0 - tmpx.0).abs() + (tmp1.1 - tmpx.1).abs() < (n1[x].0 - tmp.0)) {
                if ((tmp1.0 - tmpx.0).abs() + (tmp1.1 - tmpx.1).abs() - (n1[x].0 - tmp.0)).abs() % 2
                    == 1
                {
                    return print!("No");
                }
            }
            tmp = (n1[x].0, n1[x].1);
        } else {
            return print!("No");
        }
    }
    return print!("Yes");
}