LuoTianyi and XOR-Tree
Description
LuoTianyi 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.
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.
Print a single integer — the minimum number of operations.
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.
6
3 5 7 5 8 4
1 2
1 3
1 4
3 5
4 6
8
7 10 7 16 19 9 16 11
1 5
4 2
6 5
5 2
7 2
2 3
3 8
4
1 2 1 2
1 2
2 3
4 3
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
3
3
0
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.