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 <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long ll;
const int MAXN = 100010;
const int MAXM = 320;
int M = 316;

int n, q, a[MAXN];
int pos[MAXN], lp[MAXN], rp[MAXN], p[MAXN], tag[MAXN], cnt[MAXN];

inline int getfa(int x) {return max(1, a[x] - tag[pos[x]]);}
inline void recalc(int x)
{
	for(int i=lp[x];i<=rp[x];i++)
	{
		int fa = getfa(i);
		p[i] = x == 1 ? 1 : pos[fa] < pos[i] ? i : p[fa];
	}
}
inline void modify(int l, int r, int v)
{
	if(pos[l] == pos[r])
	{
		for(int i=l;i<=r;i++)
			a[i] = max(1, a[i] - v);
		recalc(pos[l]);
	}
	else
	{
		for(int i=l;i<=rp[pos[l]];i++)
			a[i] = max(1, a[i] - v);
		for(int i=lp[pos[r]];i<=r;i++)
			a[i] = max(1, a[i] - v);
		recalc(pos[l]);
		recalc(pos[r]);
		for(int i=pos[l]+1;i<pos[r];i++)
		{
			tag[i] = min(n, tag[i] + v);
			++cnt[i];
			if(cnt[i] <= M) recalc(i);
		}
			
	}
}
inline int query(int x, int y)
{
	while(p[x] != p[y])
	{
		if(p[x] < p[y]) swap(x, y);
		x = getfa(p[x]);
	}
	while(getfa(x) != getfa(y))
	{
		if(getfa(x) < getfa(y)) swap(x, y);
		x = getfa(x);
	}
	if(x == y) return x;
	return getfa(x);
}

int main()
{
//	freopen("data.in","r",stdin);
//	freopen("mine.out","w",stdout);
	scanf("%d%d",&n,&q);
	M = sqrt(n);
	for(int i=2;i<=n;i++) scanf("%d",a+i);
	for(int i=1;i<=n;i++)
		pos[i] = (i-1)/M+1;
	for(int i=1;i<=pos[n];i++)
		lp[i] = (i-1)*M+1, rp[i] = M*i;
	rp[pos[n]] = n;
	for(int i=1;i<=pos[n];i++) recalc(i);
	while(q--)
	{
		int opt, x, y, z;
		scanf("%d%d%d",&opt,&x,&y);
		if(opt&1)
		{
			scanf("%d",&z);
			modify(x, y, z);
		}
		else printf("%d\n",query(x, y));
	}
	return 0;
}
 	  		 		   			       	  	   		