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 ≤ a_i<i for 2 ≤ i ≤ 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 ≤ l ≤ r ≤ n, 1 ≤ x ≤ 10^5). You need to replace a_i with max(a_i-x,1) for all i with l ≤ i ≤ r.
  2. She'll give you two integers u, v (1 ≤ u, v ≤ n). You need to find the [LCA](https://en.wikipedia.org/wiki/Lowest_common_ancestor) of nodes u and v (their lowest common ancestor).

Input

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

The second line contains n-1 integers a_2, a_3,..., a_n (1 ≤ 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 ≤ l ≤ r ≤ n, 1 ≤ x ≤ 10^5), meaning that you have to replace a_i with max(a_i-x,1) for all i with l ≤ i ≤ r.

If t = 2, this represents the query of the second type. Then, two integers will follow: u and v (1 ≤ u, v ≤ 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.

Output

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

Example

Input


6 4
1 2 3 3 4
2 3 4
1 2 3 1
2 5 6
2 2 3


Output


3
3
1

Note

The tree in example is shown below.

<image>

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

<image>
#include<bits/stdc++.h>
using namespace std; 
const int N=1e5+77,B=330; 
int n,q,a[N],t[N],b[N],l[N],r[N],m,g[N]; 
int nx(int x){return max(1,a[x]-t[b[x]]); }
int jp(int x){if(t[b[x]]>=B)return nx(x); return g[x]; }
void reb(int x)
{
	for(int i=l[x]; i<=r[x]; i++)
	{
		int j=nx(i); 
		if(j<l[x])
			g[i]=j; 
		else
			g[i]=g[j]; 
	}
}
void update(int x,int y,int v)
{
	if(b[x]==b[y])
	{
		for(int i=x; i<=y; i++)
			a[i]=max(1,a[i]-v); 
		reb(b[x]); 
		return; 
	}
	for(int i=x; i<=r[b[x]]; i++)
		a[i]=max(1,a[i]-v); 
	reb(b[x]); 
	for(int i=b[x]+1; i<b[y]; i++)
	{
		t[i]=min(t[i]+v,n); 
		if(t[i]<B)
			reb(i); 
	}
	for(int i=l[b[y]]; i<=y; i++)
		a[i]=max(1,a[i]-v); 
	reb(b[y]); 
}
void lca(int u,int v)
{
	int lu=u,lv=v; 
	while(u!=v)
	{
		if(u<v)
			swap(u,v),swap(lu,lv); 
		lu=u; u=jp(u); 
	}
	u=lu,v=lv; 
	while(u!=v)
	{
		if(u<v)
			swap(u,v); 
		u=nx(u); 
	}
	printf("%d\n",u); 
}
int main()
{
	scanf("%d%d",&n,&q); 
	for(int i=2; i<=n; i++)
		scanf("%d",&a[i]); 
	for(int x=1; x<=n; x+=B)
	{
		int y=min(x+B-1,n); 
		m++; 
		l[m]=x,r[m]=y; 
		for(int i=x; i<=y; i++)
			b[i]=m; 
		reb(m); 
	}
	while(q--)
	{
		int t; 
		scanf("%d",&t); 
		if(t==1)
		{
			int x,y,v; 
			scanf("%d%d%d",&x,&y,&v); 
			update(x,y,v); 
		}
		else
		{
			int u,v; 
			scanf("%d%d",&u,&v); 
			lca(u,v); 
		}
	}

}
    		 			 	 	   					    			