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: Wrong Answer)

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <bitset>

using namespace std;
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(a) (a).begin(), (a).end()

int positive(int n){
    return (n >= 0) ? n : -n;
}

int main() {
    int N, x=0, y=0,t=0, t_now, x_now,y_now;
    cin >> N;
    REP(i, N){
        cin >> t_now >> x_now >> y_now;
        int diff_p = positive(x_now - x) + positive(y_now - y);
        int diff_t = t_now - t;

        if (diff_p > diff_t || (diff_t - diff_p) % 2 != 0){
            cout << "NO";
            return 0;
        }
        x = x_now; y=y_now; t=t_now;
    }
    cout << "YES";
    return 0;
}