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 Python (3.4.3) (judge verdict: Accepted)

import sys
n=int(input())
t=[]
x=[]
y=[]
t.append(0)
x.append(0)
y.append(0)
for i in range(1,n+1):
    a,b,c=map(int,input().split())
    t.append(a)
    x.append(b)
    y.append(c)
for j in range(1,n+1):
    S = abs(x[j]-x[j-1])+abs(y[j]-y[j-1])  #1つの移動最短距離
    T =t[j]-t[j-1] #その所用時間
    if S>T: #もし所用時間よりも距離が長ければ無理
        print("No")
        sys.exit()
    elif S%2==0: #もしも距離が偶数であれば
        if T%2==0: #時間も偶数であればOK
            continue
        else: #距離が偶数で時間が奇数ならダメ
            print("No")
            sys.exit()
    elif T%2!=0: #距離が奇数で、時間も奇数ならOK
        continue
    else:
        print("No")
        sys.exit()
print("Yes")