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)

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

typedef long long ll;
typedef long double ld;
typedef vector<vector<int>> Graph;
#define pi 3.14159265359
#define inf 2147483647
#define INF 9223372036854775807
#define mod 1000000007
#define mod2 998244353

int main() {
    int N; cin >> N;
    vector<pair<int,pair<int, int>>> T(N);
    for(int i = 0; i < N; i++) cin >> T[i].first >> T[i].second.first >> T[i].second.second;
    sort(T.begin(), T.end());
    int tim = 0;
    int prex = 0, prey = 0;
    string ans = "Yes";
    for(int i = 0; i < N; i++) {
        int nowx = T[i].second.first, nowy = T[i].second.second;
        int dist = abs(nowx - prex) + abs(nowy - prey);
        if((T[i].first - tim - dist) % 2 == 0 && T[i].first - tim - dist > 0) {
            tim += dist;
            if(tim > T[i].first) {
                ans = "No";
                break;
            }
            prex = nowx, prey = nowy;
        }
        else {
            ans = "No";
            break;
        }
    }
    cout << ans << endl;
    return 0;
}