Codeforces Round 872 (Div. 1) |
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data structures

dfs and similar

dp

dsu

greedy

trees

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C. LuoTianyi and XOR-Tree

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputLuoTianyi gives you a tree with values in its vertices, and the root of the tree is vertex $$$1$$$.

In one operation, you can change the value in one vertex to any non-negative integer.

Now you need to find the minimum number of operations you need to perform to make each path from the root to leaf$$$^{\dagger}$$$ has a bitwise XOR value of zero.

$$$^{\dagger}$$$A leaf in a rooted tree is a vertex that has exactly one neighbor and is not a root.

Input

The first line contains a single integer $$$n$$$ ($$$2 \le n \le 10^5$$$) — the number of vertices in the tree.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the $$$i$$$-th number represents the value in the $$$i$$$-th vertex.

Next $$$n−1$$$ lines describe the edges of the tree. The $$$i$$$-th line contains two integers $$$u_i$$$ and $$$v_i$$$ ($$$1 \le u_i,v_i \le n, u_i \neq v_i$$$) — the vertices connected by an edge of the tree. It's guaranteed that the given edges form a tree.

Output

Print a single integer — the minimum number of operations.

Examples

Input

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

Output

3

Input

8 7 10 7 16 19 9 16 11 1 5 4 2 6 5 5 2 7 2 2 3 3 8

Output

3

Input

4 1 2 1 2 1 2 2 3 4 3

Output

0

Input

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

Output

2

Note

The tree in the first example:

If we change the value in the vertex $$$2$$$ to $$$3$$$, the value in the vertex $$$5$$$ to $$$4$$$, and the value in the vertex $$$6$$$ to $$$6$$$, then the tree will be ok.

The bitwise XOR from the root to the leaf $$$2$$$ will be $$$3 \oplus 3=0$$$.

The bitwise XOR from the root to the leaf $$$5$$$ will be $$$4 \oplus 7 \oplus 3=0$$$.

The bitwise XOR from the root to the leaf $$$6$$$ will be $$$6 \oplus 5 \oplus 3=0$$$.

The tree in the second example:

If we change the value in the vertex $$$2$$$ to $$$4$$$, the value in the vertex $$$3$$$ to $$$27$$$, and the value in the vertex $$$6$$$ to $$$20$$$, then the tree will be ok.

The bitwise XOR from the root to the leaf $$$6$$$ will be $$$20 \oplus 19 \oplus 7=0$$$.

The bitwise XOR from the root to the leaf $$$8$$$ will be $$$11 \oplus 27 \oplus 4 \oplus 19 \oplus 7=0$$$.

The bitwise XOR from the root to the leaf $$$4$$$ will be $$$16 \oplus 4 \oplus 19 \oplus 7=0$$$.

The bitwise XOR from the root to the leaf $$$7$$$ will be $$$16 \oplus 4 \oplus 19 \oplus 7=0$$$.

In the third example, the only leaf is the vertex $$$4$$$ and the bitwise XOR on the path to it is $$$1 \oplus 2 \oplus 1 \oplus 2 = 0$$$, so we don't need to change values.

In the fourth example, we can change the value in the vertex $$$1$$$ to $$$5$$$, and the value in the vertex $$$4$$$ to $$$0$$$.

Here $$$\oplus$$$ denotes the bitwise XOR operation.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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