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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n,T,X,Y;
    cin >> n;
    vector<int> t(n+1),x(n+1),y(n+1);
    t.at(0) = 0;
    x.at(0) = 0;
    y.at(0) = 0;
    for(int i = 1;i < n+1;i ++){
        cin >> t.at(i) >> x.at(i) >> y.at(i);
    }

    for(int i = 1;i < n+1;i ++){
        if(x.at(i-1)<x.at(i)){
            X = x.at(i) - x.at(i-1);
        }else{
            X = x.at(i-1) - x.at(i);
        }

        if(y.at(i-1)<y.at(i)){
            Y = y.at(i) - y.at(i-1);
        }else{
            Y = y.at(i-1) - y.at(i);
        }

        T = t.at(i) -t.at(i-1);

        if(T >= X+Y && T%2 == (X+Y)%2){

        }else{
            cout << "No" << endl;
            return 0;
        }

    }

    cout << "Yes" << endl;

}