Problem Description

You ye Jiu yuan is the daughter of the Great GOD Emancipator. And when she becomes an adult, she will be queen of Tusikur, so she wanted to travel the world while she was still young. In a country, she found a small pub called Whitehouse. Just as she was about to go in for a drink, the boss Carola appeared. And ask her to solve this problem or she will not be allowed to enter the pub. The problem description is as follows:
There is a tree with nnn nodes, each node i contains weight a[i], the initial value of a[i] is 0. The root number of the tree is 1. Now you need to do the following operations:

1. Multiply all weight on the path from u to v by x
2. For all weight on the path from u to v, increasing x to them
3. For all weight on the path from u to v, change them to the bitwise NOT of them
4. Ask the sum of the weight on the path from u to v

The answer modulo $2^{64}$.

Jiu Yuan is a clever girl, but she was not good at algorithm, so she hopes that you can help her solve this problem. Ding∽∽∽
The bitwise NOT is a unary operation that performs logical negation on each bit, forming the ones’ complement of the given binary value. Bits that are 0 become 1, and those that are 1 become 0. For example:

NOT 0111 (decimal 7) = 1000 (decimal 8)

NOT 10101011 = 01010100

Input

The input contains multiple groups of data.
For each group of data, the first line contains a number of n, and the number of nodes.
The second line contains (n−1) integers bi, which means that the father node of node (i+1) is bi.
The third line contains one integer m, which means the number of operations，
The next m lines contain the following four operations:
At first, we input one integer opt

1. If opt is 1, then input 3 integers, u,v,x, which means multiply all weight on the path from u to v by x
2. If opt is 2, then input 3 integers, u,v,x, which means for all weight on the path from u to v, increasing x to them
3. If opt is 3, then input 2 integers, u,v, which means for all weight on the path from u to v, change them to the bitwise NOT of them
4. If opt is 4, then input 2 integers, u,v, and ask the sum of the weights on the path from u to v

$1≤n,m,u,v≤10^5,1≤x<2^{64}$

Output

For each operation 4, output the answer.

Examples

Input
7
1 1 1 2 2 4
5
2 5 6 1
1 1 6 2
4 5 6
3 5 2
4 2 2
2
1
4
3 1 2
4 1 2
3 1 1
4 1 1

Output
5
18446744073709551613
18446744073709551614
0

【題目連結】 Jiu Yuan Wants to Eat
【題意】
給定一棵以1為根的樹，存在路徑加，路徑乘，路徑的節點權值取反，查詢路徑和等四個操作。
【思路】
路徑加、路徑乘、路徑查詢是最基本的樹鏈剖分操作，但是這裡還涉及到路徑取反操作。11000 取反為00111，相當於(11111-11000)=00111即將原來的值先乘上-1，再加$2^{64}-1$,相當於減1。每次膜$2^{64}$，利用unsigned long long 自然溢位即可

這樣就可以把取反操作轉化為加法和乘法操作。

對於路徑加和乘法的計算，可以使用兩個陣列add, mul來線段樹維護，線段樹上每個節點的值為$tree[]*mul+add$

修改時：
$+val，則add+val$
$*val，則add*val，mul*val$

我是程式碼