CF1491H.Yuezheng Ling and Dynamic Tree

Yuezheng Ling gives Luo Tianyi a tree which has $n$ nodes, rooted at $1$ .

Luo Tianyi will tell you that the parent of the $i$ -th node is $a_i$ ( $1 \leq a_i<i$ for $2 \le i \le n$ ), and she will ask you to perform $q$ queries of $2$ types:

1. She'll give you three integers $l$ , $r$ and $x$ ( $2 \le l \le r \le n$ , $1 \le x \le 10^5$ ). You need to replace $a_i$ with $\max(a_i-x,1)$ for all $i$ with $l \leq i \leq r$ .
2. She'll give you two integers $u$ , $v$ ( $1 \le u, v \le n$ ). You need to find the LCA of nodes $u$ and $v$ (their lowest common ancestor).

The first line contains two integers $n$ and $q$ ( $2\leq n,q \leq 10^5$ ) — the number of nodes and the number of queries, respectively.

The second line contains $n-1$ integers $a_2, a_3,\dots, a_n$ ( $1 \le a_i < i$ ), where $a_i$ is the parent of the node $i$ .

Next $q$ lines contain queries. For each query, the first integer of each line is $t$ ( $t = 1$ or $2$ ) — the type of the query.

If $t = 1$ , this represents the query of the first type. Then, three integers will follow: $l$ , $r$ , $x$ ( $2 \le l \le r \le n$ , $1 \le x \le 10^5$ ), meaning that you have to replace $a_i$ with $\max(a_i-x,1)$ for all $i$ with $l \leq i \leq r$ .

If $t = 2$ , this represents the query of the second type. Then, two integers will follow: $u$ and $v$ ( $1 \le u, v \le n$ ), and you have to find the LCA of $u$ and $v$ .

It's guaranteed that there is at least one query of the second type.

For each query of the second type output answer on a new line.

The tree in example is shown below.

After the query of the first type, the tree changes and is looking as shown below.