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 (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n: usize = read();
    let mut t: Vec<i32> = vec![0; n + 1];
    let mut x: Vec<i32> = vec![0; n + 1];
    let mut y: Vec<i32> = vec![0; n + 1];

    t[0] = 0;
    x[0] = 0;
    y[0] = 0;
    for i in 1..n+1 {
        t[i] = read();
        x[i] = read();
        y[i] = read();
    }

    let mut succeeded = true;

    // その場に留まることができないことが考慮に入っていない
    for i in 0..n {
        if (t[i+1] - t[i]) < (x[i+1] - x[i]).abs() + (y[i+1] - y[i]).abs()
            || (t[i+1] - t[i]) % 2 != ((x[i+1] - x[i]).abs() + (y[i+1] - y[i]).abs()) % 2 {
            succeeded = false;
            break;
        }
    }

    println!("{}", if succeeded { "Yes" } else { "No" });
}