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 _CRT_SECURE_NO_WARNINGS
#include"bits/stdc++.h"
#ifdef _DEBUG
#define DBG(n) n
#else
#define DBG(n)
#endif
#define INF         1e9
#define INFLL       1e18
#define EPS         1e-9
#define REP(i,n)    for(ll i=0,i##_len=(n);i<i##_len;++i)
#define REP1(i,n)   for(ll i=1,i##_len=(n);i<=i##_len;++i)
#define REPR(i,n)   for(ll i=(n)-1;i>=0;--i)
#define REPR1(i,n)  for(ll i=(n);i>0;--i)
#define REPC(i,obj) for(auto i:obj)
#define ALL(obj)    (obj).begin(),(obj).end()
#define SETP(n)     cout << fixed << setprecision(n)
using namespace std;
using ll = long long;
template<typename T = ll>inline T in() { T ret; cin >> ret; return ret; }

signed main() {
	int n = in();
	vector<array<int, 3>>cood(n + 1);
	cood[0] = { 0, 0, 0 };
	REP1(i, n) {
		int t = in(), x = in(), y = in();
		cood[i] = { t, x, y };
	}
	REP(i, n) {
		int dt = cood[i + 1][0] - cood[i][0];
		int manhattan = abs(cood[i + 1][1] - cood[i][1]) + abs(cood[i + 1][2] - cood[i][2]);
		if (manhattan > dt || (dt - manhattan) % 2 != 0) {
			cout << "No" << endl;
			return 0;
		}
	}
	cout << "Yes" << endl;
}