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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 10e17
#define rep(i,n) for(long long i=0; i<n; i++)
#define repr(i,n,m) for(long long i=m; i<n; i++)
#define mod 1000000007
#define sorti(x) sort(x.begin(), x.end())
#define sortd(x) sort(x.begin(), x.end(), std::greater<long long>())
#define debug(x) std::cerr << (x) << std::endl;
#define roll(x) for (auto&& itr : x) { cerr << (itr) << " "; }

template <class T> inline void chmax(T &ans, T t) { if (t > ans) ans = t;}
template <class T> inline void chmin(T &ans, T t) { if (t < ans) ans = t;}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N;
  cin >> N;
  vector<int> X(N), Y(N), T(N);
  repr(i, N, 0) {
    cin >> T[i] >> X[i] >> Y[i];
  }

  int x,y,time;
  x = 0, y = 0, time = 0;
  for (int i = 0; i < N; ++i) {
    int nx,ny,dist,rtime;
    nx = X[i], ny = Y[i], rtime = T[i] - time;
    dist = max(abs(x - X[i]) + abs(Y[i] - y), 0);
    if (dist > rtime || rtime % dist == 1) {
      cout << "No" << endl;
      return 0;
    }

    x = nx, y = ny, time = rtime;
  }

  cout << "Yes" << endl;
}