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: Accepted)

#include <iostream>

using namespace std;

bool is_ok(int t_diff, int d);

int main() {
  int n;
  cin >> n;
  int t[100001];
  int x[100001];
  int y[100001];

  t[0] = x[0] = y[0] = 0;
  // 入力
  for (int i = 1; i <= n; i++) {
    cin >> t[i] >> x[i] >> y[i];

  }
  // 確認
  for (int i = 0; i <= n-1; i++) {
    int t_diff, x_diff, y_diff;
    t_diff = t[i+1] - t[i];
    x_diff = x[i+1] - x[i];
    y_diff = y[i+1] - y[i];
    int d = abs(x_diff) + abs(y_diff);
    //cout << i << "," << t[i+1] << "," << t[i] << "," << t_diff << "," << d << "----" << endl;
    if (!is_ok(t_diff, d)) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
}

bool is_ok(int t_diff, int d) {
  if (d > t_diff) {
    return false;
  }
  if (d % 2 != t_diff % 2) {
    return false;
  }
  return true;
}