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<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=100005;
const int M=350;
int l[M],r[M],f[M],pos[N],cnt[M],a[N],b[N],k,p;
void upd(int x){
	for(int i=l[x];i<=r[x];i++) a[i]=max(1,a[i]-f[x]);
	f[x]=0;
	for(int i=l[x];i<=r[x];i++){
		if(a[i]<l[x]) b[i]=a[i];
		else b[i]=b[a[i]];
	} 
}
void change(int x,int y,int z){
	int L=pos[x],R=pos[y];
	if(L==R){
		for(int i=x;i<=y;i++) a[i]=max(a[i]-z,1);
		upd(L);return;
	}
	for(int i=x;i<=r[L];i++) a[i]=max(a[i]-z,1);upd(L);
	for(int i=l[R];i<=y;i++) a[i]=max(a[i]-z,1);upd(R);
	for(int i=L+1;i<R;i++){
		cnt[i]++;f[i]=min(N,f[i]+z);
		if(cnt[i]<=k) upd(i); 
	}
}
int gb(int x){return max(1,b[x]-f[pos[x]]);}
int lca(int x,int y){
	while(1){
		if(pos[x]<pos[y]) swap(x,y);
		if(pos[x]!=pos[y]) x=gb(x);
		else{
			if(gb(x)!=gb(y)) x=gb(x),y=gb(y);
			else break;
		}
	}
	while(x!=y){
		if(x<y) swap(x,y);
		x=max(1,a[x]-f[pos[x]]);
	}
	return x;
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	k=sqrt(n),p=(n-1)/k+1;
	for(int i=2;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<=n;i++) pos[i]=((i-1)/k+1);
	for(int i=1;i<=p;i++) l[i]=(i-1)*k+1,r[i]=min(k*i,n);
	for(int i=1;i<=p;i++) upd(i);
	for(int i=1;i<=m;i++){
		int opt;
		scanf("%d",&opt);
		if(opt==1){
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			change(x,y,z);
		}
		if(opt==2){
			int x,y;
			scanf("%d%d",&x,&y);
			printf("%d\n",lca(x,y));
		}
	}
 } 
	  	 						 				  		 				 		 		