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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define leading zero str.erase(0, min(str.find_first_not_of('0'), str.size()-1));
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
string text="abcdefghijklmnopqrstuvwxyz";
const int maxn=1e6+7;
//        .--------------.
//        | Try First One|
//        '--------------'
//                |     .--------------.
//                |     |              |
//                V     V              |
//              .--------------.       |
//              |      AC.     |<---.  |
//              '--------------'    |  |
//              (True)|  |(False)   |  |
//           .--------'  |          |  |
//           |           V          |  |
//           |  .--------------.    |  |
//           |  |   Try Again  |----'  |
//           |  '--------------'       |
//           |                         |
//           |  .--------------.       |
//           '->| Try Next One |-------'
//              '--------------'
ll bin_pow(ll a,ll b,ll m)
{
    ll res=1;
    a%=m;
    while(b>0)
    {
        if(b&1)
            res=res*a%m;
        b>>=1;
        a=a*a%m;
    }
    return res;
}
bool miller_rabin(ll d,ll n)
{
    ll a=2+rand()%(n-4);
    ll x=bin_pow(a,d,n);
    if(x==1 || x==n-1)
        return true;
    while(d!=n-1)
    {
        x=(x*x)%n;
        d*=2;
        if(x==1)
            return false;
        if(x==n-1)
            return true;
    }
    return false;
}
bool prime(ll n,ll k)
{
    if(n==1 || n==4)
        return false;
    if(n<=3)
        return true;
    ll d=n-1;
    while(d%2==0)
        d/=2;
    for(int i=0; i<k; i++)
    {
        if(!miller_rabin(d,n))
            return false;
    }
    return true;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin>>n;
    bool ok=true;
    int k,dif1,dif2,dif;
    for(int i=0;i<n;i++){
        int t,x,y;
        cin>>t>>x>>y;
        if(i==0){
            k=t;
            dif1=x;
            dif2=y;
        }
        else{
            k=abs(k-t);
            dif=abs(dif1-x)+abs(dif2-y);
            dif1=x;
            dif2=y;
            if(k%2!=dif%2)ok=false;
        }
    }
    if(n!=1 && ok)puts("Yes");
    else puts("No");
}