Problem: 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>

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
int const B=317;
int n,m,a[100010],tag[100010],L[320],R[320],cnt[320],b[100010],bel[100010];
int to(int const &x){if(x==1)return 0;return std::max(b[x]-tag[bel[x]],1);}
int dd(int const &x){if(x==1)return 0;return std::max(a[x]-tag[bel[x]],1);}
void part(int const &l,int const &r,int const &x){
	int bl=bel[l];
	if(cnt[bl]>B){
		for(int i=l;i<=r;i++)a[i]=std::max(a[i]-x,1);
		for(int i=l;i<=r;i++)b[i]=a[i];
		return;
	}
	for(int i=l;i<=r;i++)a[i]=std::max(a[i]-x,1);
	for(int i=l;i<=r;i++)b[i]=a[i];
	for(int i=L[bl];i<=R[bl];i++)if(a[i]<L[bl])b[i]=a[i];else b[i]=b[a[i]];
}
void whole(int const &bl,int const &x){
	if(cnt[bl]>B)return tag[bl]=std::min(n,x+tag[bl]),void();
	++cnt[bl];
	for(int i=L[bl];i<=R[bl];i++)a[i]=std::max(a[i]-x,1);
	for(int i=L[bl];i<=R[bl];i++)if(a[i]<L[bl])b[i]=a[i];else b[i]=b[a[i]];
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=2;i<=n;i++)scanf("%d",a+i),bel[i]=(i-1)/B+1;
	for(int i=1;i<=n;i++)R[bel[i]]=i;
	for(int i=n;i;i--)L[bel[i]]=i;
	for(int bl=1;bl<=bel[n];bl++)
		for(int i=L[bl];i<=R[bl];i++)if(a[i]<L[bl])b[i]=a[i];else b[i]=b[a[i]];
	while(m--){
		int op;
		scanf("%d",&op);
		if(op==1){
			int l,r,x;
			scanf("%d%d%d",&l,&r,&x);
			if(bel[l]==bel[r]){
				part(l,r,x);
			}else{
				part(l,R[bel[l]],x);
				part(L[bel[r]],r,x);
				for(int i=bel[l]+1;i<bel[r];i++)whole(i,x);
			}
		}else{
			int u,v;
			scanf("%d%d",&u,&v);
			while(to(u)!=to(v))
				if(to(u)>=to(v)) u=to(u);
				else v=to(v);
			while(u!=v)
				if(dd(u)>=dd(v)) u=dd(u);
				else v=dd(v);
			printf("%d\n",u);
		}
	}
	return 0;
}