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<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
#define REP(i, limit) for(int i=0; i< limit; ++i)
#define FOR(i, j, limit) for(int i=j; i< limit; ++i)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ECHO(x) cout<<x<<endl;
#define LL long long
int INF = 2147483647;
LL LINF = 9223372036854775807;
LL MOD = 1000000007;
typedef pair<int, int>P;
typedef pair<LL, LL>PL;
typedef vector<int> vi;
typedef vector<LL> vl;


int main(){
  int n;
  cin>>n;
  int t[n+1], x[n+1], y[n+1];
  t[0]=0;
  x[0]=0;
  y[0]=0;
  REP(i, n){
    cin>>t[i+1]>>x[i+1]>>y[i+1];
  }
  bool ans=true;
  REP(i, n){
    if(t[i+1]-t[i]<abs(x[i+1]-x[i]) + abs(y[i+1]-y[i])) {
      ans=false;
      break;
    }
    if(abs(t[i+1]-t[i]-abs(x[i+1]-x[i])-abs(y[i+1]-y[i]))%2) {
      ans=false;
      break;
    }
  }

  if(ans) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;


  return 0;
}