Codeforces Round #396(Div. 2)E. Mahmoud and a xor trip【二进制拆位+Dfs处理】

E. Mahmoud and a xor trip

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Mahmoud and Ehab live in a country with n cities numbered from
1 to n and connected by
n - 1 undirected roads. It's guaranteed that you can reach any city from any other using these roads. Each city has a number
ai attached to it.

We define the distance from city x to city
y as the xor of numbers attached to the cities on the path from
x to y
(including both x and
y). In other words if values attached to the cities on the path from
x to y form an array
p of length l then the distance between them is


, where


is bitwise
xor operation.

Mahmoud and Ehab want to choose two cities and make a journey from one to another. The index of the start city is always less than or equal to the index of the finish city (they may start and finish in the same city and in this case the distance equals the
number attached to that city). They can't determine the two cities so they try every city as a start and every city with greater index as a finish. They want to know the total distance between all pairs of cities.

Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of cities in Mahmoud and Ehab's country.

Then the second line contains n integers
a1, a2, ..., an (0 ≤ ai ≤ 106)
which represent the numbers attached to the cities. Integer
ai is attached to the city
i.

Each of the next n  -  1 lines contains two integers
u and v (1  ≤  u,  v  ≤  n,
u  ≠  v), denoting that there is an undirected road between cities
u and v. It's guaranteed that you can reach any city from any other using these roads.

Output
Output one number denoting the total distance between all pairs of cities.

Examples

Input

3
1 2 3
1 2
2 3

Output

10

Input

5
1 2 3 4 5
1 2
2 3
3 4
3 5

Output

52

Input

5
10 9 8 7 6
1 2
2 3
3 4
3 5

Output

131

Note
A bitwise xor takes two bit integers of equal length and performs the logical
xor operation on each pair of corresponding bits. The result in each position is
1 if only the first bit is 1 or only the second bit is
1, but will be 0 if both are
0 or both are 1. You can read more about bitwise
xor operation here: https://en.wikipedia.org/wiki/Bitwise_operation#XOR.
In the first sample the available paths are:

city 1 to itself with a distance of
1,
city 2 to itself with a distance of
2,
city 3 to itself with a distance of
3,
city 1 to city 2 with a distance of


,

city 1 to city 3 with a distance of


,

city 2 to city 3 with a distance of


.

The total distance between all pairs of cities equals 1 + 2 + 3 + 3 + 0 + 1 = 10.

题目大意：

给你N个点的一棵树，让你求每两点间距离和。

这里设定每两点之间距离为：两点间路上所有点的异或值。

思路（思路源自：http://blog.csdn.net/sjtsjt709/article/details/54926998）：

1、首先考虑直接处理O（n^2）是肯定不行的，那么接下来考虑到二进制。

对于一个二进制数来讲，每一位上的二进制操作都是相互独立的。

那么我们考虑将每一位都分开来处理。

2、那么对于每一位来讲，其贡献的价值就是异或值为1的路径数*（1<<第几位）；

那么我们接下来的任务就是Dfs处理出按照这位赋值的情况下，异或值为1的路径总数。

那么我们可以设定cnt【u】【0】表示以u为根节点，异或值为0的路径数，cnt【u】【1】表示以u为根节点，异或值为1的路径数。

那么对于每一次处理完以子节点v为根的子树之后，那么这部分有异或值为1的路径数为：cnt【u】【0】*cnt【v】【1】+cnt【u】【1】*cnt【v】【0】；

过程维护cnt【】【】数组即可。

然后将每一位的贡献值都累加起来就行啦。

Ac代码：