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>
#define get(x) max((f[x]-tag[bel[x]]),1)
#define getf(x) max((a[x]-tag[bel[x]]),1)
using namespace std;
int const B=330;
int a[100005],L[100005/B+5],R[100005/B+5],f[100005],bel[100005],tag[100005],times[100005/B+5];
void build(int x)
{
    if(tag[x]){for(int i=L[x];i<=R[x];i++)a[i]=max(a[i]-tag[x],1);tag[x]=0;}
    for(int i=L[x];i<=R[x];i++)f[i]=(a[i]<L[x])?a[i]:f[a[i]];
}
int query(int x,int y)
{
    while(get(x)!=get(y))
    {
        if(bel[x]!=bel[y])
        {
            if(bel[x]<bel[y])swap(x,y);
            x=get(x);
        }
        else x=get(x),y=get(y);
    }
    while(x!=y){if(x<y)swap(x,y);x=getf(x);}
    return x;
}
int main()
{
    int n,q,cnt=0,op,l,r,x,u,v;
    scanf("%d%d",&n,&q);
    for(int i=2;i<=n;i++)scanf("%d",&a[i]);
    while(1)
    {
        cnt++;
        L[cnt]=(cnt-1)*B+1;
        R[cnt]=min(n,cnt*B);
        for(int i=L[cnt];i<=R[cnt];i++)bel[i]=cnt;
        if(R[cnt]==n)break;
    }
    for(int i=1;i<=cnt;i++)build(i);
    while(q--)
    {
        scanf("%d",&op);
        if(op==1)
        {
            scanf("%d%d%d",&l,&r,&x);
            if(bel[l]==bel[r])
            {
                for(int i=l;i<=r;i++)a[i]=max(a[i]-x,1);
                build(bel[l]);
            }
            else
            {
                for(int i=l;i<=R[bel[l]];i++)a[i]=max(a[i]-x,1);
                build(bel[l]);
                for(int i=L[bel[r]];i<=r;i++)a[i]=max(a[i]-x,1);
                build(bel[r]);
                for(int i=bel[l]+1;i<bel[r];i++)
                {
                    if(times[i]>=B){tag[i]=min(tag[i]+x,n);continue;}
                    times[i]+=x;
                    for(int j=L[i];j<=R[i];j++)a[j]=max(a[j]-x,1);
                    build(i);
                }
            }
        }
        else scanf("%d%d",&u,&v),printf("%d\n",query(u,v));
    }
    return 0;
}