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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>

using namespace std;

int main()
{
        int i;
        int n;

        vector<int> T;
        vector<int> X;
        vector<int> Y;

        int a, a1, a2, b, c;
        int s_x, s_y = 0;
        int min;

        cin >> n;

        for (i=0; i<n; i++) {
                cin >> a;
                cin >> b;
                cin >> c;
                T.push_back(a);
                X.push_back(b);
                Y.push_back(c);

        }


        a1 = 0;

        for (i=0; i<T.size(); i++) {

                a2 = T.back();
                b = X.back();
                c = Y.back();

                min = (b - s_x) + (c - s_y);

                if ((a2 - a1) < min) {
                        break;
                }

                // (a2-a1) % 2 == 0 && min % 2 == 0
                // (a2-a1) % 2 == 1 && min % 2 == 1
                if (((a2-a1) % 2 == 0) && (min % 2 == 1)) {
                        break;
                }

                if (((a2-a1) % 2 == 1) && (min % 2 == 0)) {
                        break;
                }
                a1 = a2;
                s_x = b;
                s_y = c;

        }

        if (i==T.size()) {
                cout << "Yes" << endl;
        }else{
                cout << "No" << endl;
        };
}