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 il inline
#define ll long long
const int N=2e5+5,B=316;

int n,q,b[N],c[N]; ll a[N],lz[N];

il int jump(int x){return x?std::max(a[x]-lz[x/B],0ll):-1;}

il void rebuild(int x,int v)
{
	if (c[x]+v>B) return ; c[x]+=v; int i;
	for (i=x*B; i<x*B+B; i++) b[i]=jump(i)<x*B?i:b[jump(i)];
}

il void add(int x,int y,int v)
{
	int i; for (i=x/B*B; i<x; i++) a[i]+=v;
	for (i=y+1; i<y/B*B+B; i++) a[i]+=v;
	for (i=x/B; i<=y/B; i++) lz[i]+=v,rebuild(i,i!=x/B&&i!=y/B);
}

il int LCA(int x,int y)
{
	for ( ; b[x]!=b[y]; x=jump(b[x])) if (jump(b[x])<jump(b[y])) std::swap(x,y);
	for ( ; x!=y; x=jump(x)) if (x<y) std::swap(x,y); return x;
}

int main()
{
	scanf("%d%d",&n,&q),a[0]=-1; int i,x,y,k;
	for (i=1; i<n; i++) scanf("%d",a+i),a[i]--;
	
	for (add(0,n-1,0); q; q--)
		if (scanf("%d%d%d",&i,&x,&y),i>1) printf("%d\n",LCA(x-1,y-1)+1);
		else scanf("%d",&k),add(x-1,y-1,k);
	
	return 0;
}