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 _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef tuple<int,int, int> tup;
const int INF = 1e9;
const int MOD = 1000000007;
const double pi = 3.141592653589793238;

ll gcd(ll a, ll b) { return __gcd(a,b); } //最大公約数
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; } //最大公倍数

void printV(vector<int> &v);
void printVV(vector<vector<int>> &vv);


int main() {
  int N;
  cin >> N;

  vector<int> t(N+1);
  vector<int> x(N+1);
  vector<int> y(N+1);
  vector<int> dt(N+1);
  vector<int> di(N+1);

  t.at(0) = 0;
  x.at(0) = 0;
  y.at(0) = 0;
  dt.at(0) = 0;
  di.at(0) = 0;
  for(int i = 1; i <= N; i++) {
    cin >> t.at(i) >> x.at(i) >> y.at(i);
    dt.at(i) = t.at(i) - t.at(i-1); // 時刻の差
    di.at(i) = abs(x.at(i) - x.at(i-1)) + abs(y.at(i) - y.at(i-1)); // 変位の合計
  }

  // printV(dt);
  // printV(di);

  for(int i = 1; i <= N; i++) {
    if(dt.at(i) - di.at(i) < 0 || (dt.at(i) - di.at(i)) % 2 == 1) { // 秒数 - 変位の合計 が 0未満
      cout << "No" << endl;
      return 0;
    }
  }

  cout << "Yes" << endl;

  return 0;
}


// ベクターの中身を全部出力
void printV(vector<int> &v) {
  cout << "  [ ";
  for(int i = 0; i < (int)v.size(); i++) {
    cout << v.at(i) << " ";
  }
  cout << "]" << endl;
}

void printVV(vector<vector<int>> &vv) {
  cout << "{" << endl;
  for(int i = 0; i < (int)vv.size(); i++) {
    cout << "  [ ";
    for(int j = 0; j < (int)vv.at(i).size(); j++) {
      cout << vv.at(i).at(j) << " ";
    }
    cout << "]," << endl;
  }
  cout << "}" << endl;
}