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)


#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
#include <map>
#include <iomanip>	
#include <deque>
#include <set>
#define rep(i,a,b) for(int i=a;i<b;++i)
#define rrep(i,a,b) for(int i=a;i>=b;--i)
#define fore(i,a) for(auto &i:a)
#define INF INT_MAX/2;
typedef long long ll;
using namespace std;
using vi = vector<int>;
using vll = vector<ll>;
ll mod = 1e9 + 7;
//ll llmax = 10000000000000000000ll;
using namespace std;
using Graph = vector<vector<int>>;


int cnt_digit(ll N)
{
	int digit = 0;

	while (N > 0)
	{
		N /= 10;
		digit++;
	}
	return digit;
}


 // 最大公約数計算
 
ll gcd(ll a, ll b)
{
	if (b == 0) {
		return a;
	}
	else {
		return gcd(b, a % b);
	}
}
// 最小公倍数の計算

ll lcm(ll a, ll b)
{
	ll g = gcd(a, b);

	return a / g * b;
}

struct union_find
{
	vector<int> par, r;
	union_find(int n)
	{
		par.resize(n);
		r.resize(n);
		init(n);
	}


	void init(int n)
	{
		for (int i = 0; i < n; i++) par[i] = i;
		for (int i = 0; i < n; i++) r[i] = 0;
	}

	int find(int x)
	{
		if (par[x] == x)return x;
		else return find(par[x]);
	}

	void unite(int x, int y)
	{
		x = find(x);
		y = find(y);

		if (r[x] < r[y])
		{
			par[x] = y;
		}
		else
		{
			par[y] = x;
			if (r[x] == r[y])
			{
				r[x]++;
			}
		}
	}
};



int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
#define N 1000;
	int n;
	cin >> n;
	vector<int> x(110000);
	vector<int> y(110000);
	vector<int> t(110000);
	t[0] = 0;
	x[0] = 0;
	y[0] = 0;
	rep(i, 1, n+1) {
		cin >> t[i] >> x[i] >> y[i];
	}

	bool ans = true;
	rep(i, 0, n) {
		int dt = t[i + 1] - t[i];
		int dist = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i]);
		if (dt < dist)ans = false;
		if (dt % 2 != dist % 2)ans = false;

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


	
	return 0;
}