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;
int n,q;
const int maxn=101000;
int f[maxn],g[maxn],tag[maxn];
int cnt[maxn],L[maxn],bel[maxn],T;
void Restore(int x){
	for(int i=L[x];i<L[x+1];++i)
		f[i]=max(f[i]-tag[x],1);
	tag[x]=0;
    for(int i=L[x];i<L[x+1];++i){
        if(f[i]<L[x]||i==1)g[i]=f[i];
        else g[i]=g[f[i]];
    }
}
void modify(int l,int r,int x){
    if(bel[l]==bel[r]){
        for(int i=l;i<=r;++i)
            f[i]=max(f[i]-x,1);
        Restore(bel[l]);
    }
    else{
        for(int i=l;i<L[bel[l]+1];++i)
            f[i]=max(f[i]-x,1);
        Restore(bel[l]);
        for(int i=L[bel[r]];i<=r;++i)
            f[i]=max(f[i]-x,1);
        Restore(bel[r]);
        for(int i=bel[l]+1;i<bel[r];++i){
            ++cnt[i];
            tag[i]=min(tag[i]+x,n);
            if(cnt[i]<=T)Restore(i);
        }
    }
}
int query(int x,int y){
    while(1){
    	if(bel[x]<bel[y])swap(x,y);
    	if(bel[x]!=bel[y])x=max(1,g[x]-tag[bel[x]]);
    	else{
    		if(max(1,g[x]-tag[bel[x]])==max(1,g[y]-tag[bel[y]]))
    			break;
    		x=max(1,g[x]-tag[bel[x]]);
    		y=max(1,g[y]-tag[bel[y]]);
    	}
    }
    while(x!=y){
    	if(x<y)swap(x,y);
    	x=max(1,f[x]-tag[bel[x]]);
    }
    return x;
}
int main(){
    scanf("%d%d",&n,&q);
    f[1]=1;
    for(int i=2;i<=n;++i)
        scanf("%d",f+i);
    T=sqrt(n);
    L[1]=1;
    for(int j=1;L[j]<=n;++j){
        L[j+1]=min(L[j]+T,n+1);
        for(int k=L[j];k<L[j+1];++k)
            bel[k]=j;
        Restore(j);
    }
    for(int i=1,op,x,y,z;i<=q;++i){
        scanf("%d",&op);
        if(op==1){
            scanf("%d%d%d",&x,&y,&z);
            modify(x,y,z);
        }
        else{
            scanf("%d%d",&x,&y);
            printf("%d\n",query(x,y));
        }
    }
    return 0;
}