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: Wrong Answer)

fn main() {
    let mut n = String::new();
    std::io::stdin().read_line(&mut n).unwrap();
    let n: u32 = n.trim().parse().unwrap();
    let mut i = 0;
    let mut txy = Vec::new();
    while ( i < n ) {
        let mut input = String::new();
        std::io::stdin().read_line(&mut input).unwrap();
        let tmp: Vec<i32> = input.trim().split_whitespace()
            .map(|x| x.parse().unwrap()).collect();
        txy.push(tmp);
        i += 1;
    }
    let mut flag = 0;
    let mut ans = 0;
    let mut i = 0;
    while ( i < n ) {
        let mut d = 0;
        let mut j = 0;
        if ( i == 0 ) {
            d = txy[i as usize][1].abs() + txy[i as usize][2].abs();
            j = txy[i as usize][0];
        } else {
            d = (txy[i as usize][1] - txy[(i - 1) as usize][1]).abs() + (txy[i as usize][2] - txy[(i - 1) as usize][2]).abs();
            j = txy[i as usize][0] - txy[(i - 1) as usize][0];
        }
        if ( j < d) {
            break;
        }
        if ( j % 2 == 1 ) {
            if ( d % 2 == 1 && flag != 1 ) {
                ans = 1;
            } else {
                flag = 1;
            }
        } else if ( j % 2 == 0 ){
            if ( d % 2 == 0 && flag != 1 ) {
                ans = 1;
            } else {
                flag = 1;
            }
        }
        i += 1;
    }
    if ( ans == 1 ) {
        println!("Yes");
    } else {
        println!("No");
    }
}