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>
#define rep(i,j,n) for(int i=(j);i<(n);i++)
#define per(i,n,j) for(int i=(n)-1;i>=(j);i--)
using ll=long long;
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const ll INF=1LL<<60;

ll n;
vector<ll> t(100100),x(100100),y(100100);

int main(){
    cin>>n;
    rep(i,0,n) cin>>t[i]>>x[i]>>y[i];

    bool flag=true;
    rep(i,0,n){
        if(t[i]<x[i]+y[i]){
            flag=false;
            break;
        }
        if(x[i]==0 && y[i]==0){
            if(t[i]%2!=0){
                flag=false;
                break;
            }
        }
        else if(t[i]%(x[i]+y[i])!=0){
            flag=false;
            break;
        }
        if(i>0){
            if(t[i]-t[i-1]<=abs(x[i]-x[i-1])+abs(y[i]-y[i-1])){
                flag=false;
                break;
            }
        }
    }
    if(flag) cout<<"Yes"<<"\n";
    else cout<<"No"<<"\n";
    return 0;
}