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 <iomanip>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <functional>
#include <limits.h>
#include <set>
#include <map>
#include <tuple>
using namespace std;

#define ll long long
#define ull unsigned long long
#define rep(i,N) for (int i=0;i<N;i++)
#define loop(i,N,M) for(int i=N;i<M;i++)
#define MAX(v) *max_element(v.begin(),v.end())
#define MIN(v) *min_element(v.begin(),v.end())
#define SORTL2S(type,v) sort(v.begin(),v.end(),greater<type>())
#define SORTS2L(type,v) sort(v.begin(),v.end())
#define SORTMF(v,func) sort(v.begin(),v.end(),func)

//変数
ll N;
vector<pair<ll, pair<ll, ll>>> vec;

//解法
void solve() {
	
	bool flag = true;

	rep(i, N) {
		ll len = vec[i].second.first + vec[i].second.second - vec[i + 1].second.first - vec[i + 1].second.second;
		len = abs(len);
		ll time = vec[i + 1].first - vec[i].first;

		if (len <= time
			&& (time - len) % 2 == 0) {
		}
		else {
			flag = false;
		}
	}

	cout << (flag ? "Yes" : "No");

	return;
}

int main() {

	//入力
	cin >> N;
	vec.resize(N+1);

	vec[0].first = 0;
	vec[0].second.first = 0;
	vec[0].second.second = 0;

	rep(i, N) cin >> vec[i+1].first >> vec[i+1].second.first >> vec[i+1].second.second;

	solve();
	cout << endl;
	return 0;
}