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;
typedef long double ld;
const long long INF = 1e18;
# define len(x) ((int)(x).size())
# define rep(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
# define reps(i, n) for(int i=1, i##_len=(n); i<=i##_len; ++i)
# define rrep(i, n) for(int i=((int)(n)-1); i>=0; --i)
# define rreps(i, n) for(int i=((int)(n)); i>0; --i)
# define foreps(i, m, n) for(int i = m;i < n;i++)
# define ALL(x) (x).begin(), (x).end()
# define rall(x) (x).rbegin(), (x).rend()
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int main() {
    ll N; cin >> N;
    vector<ll> T(N), X(N), Y(N);
    rep(i, N) {
        cin >> T[i] >> X[i] >> Y[i];
    } 

    bool ans = true;
    for (ll i = 0; i < N; i++) {
        ll dt = abs(T[i + 1] - T[i]), distance = abs(X[i + 1] - X[i]) + abs(Y[i + 1] - Y[i]);
        if (dt < distance) ans = false;
        if (dt % 2 != distance % 2) ans = false;
    }
    cout << (ans ? "Yes" : "No") << endl;
}