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.