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,i,j;
  bool ok=true;
  cin>>N;
  vector<vector<int>> time(N,vector<int>(3,0));
  for(i=0;i<N;i++){
    for(j=0;j<3;j++){
    cin>>time.at(i).at(j);
    }
  }

  int s_x=0;
  int s_y=0;
  int s_tt=0;
  for (int k=0;k<N;k++){
    
  int x=time.at(k).at(1);
  int y=time.at(k).at(2);
  int t=time.at(k).at(0);
  if((abs(x-s_x)+abs(y-s_y))%2==abs(t-s_tt)%2 && (abs(x-s_x)+abs(y-s_y))<=(t-s_tt)){
    s_x=x;
    s_y=y;
    s_tt=t;
  }else{
    ok=false;
   break;
  }
  
  }
if(ok==false){
 cout<<"No"<<endl; 
}else{
 cout<<"Yes"<<endl; 
}
}