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<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
int qmax(int a, int b){
	return (a > b) ? a : b;
}
int qmin(int a, int b){
	return (a < b) ? a : b;
}
const int BS = 340;

int n, q;
int a[100400];
bool has_to[444];
int offset[444];
int go[100400];
void rebuild(int bn){
	has_to[bn] = 0;
	for(int i = bn * BS; i < (bn + 1) * BS; i++){
		if(a[i] < bn * BS) go[i] = a[i];
		else go[i] = go[a[i]], has_to[bn] = 1;
	}
}
void go_block(int &now){
	if(has_to[now / BS]) now = go[now];
	else now = qmax(0, a[now] - offset[now / BS]);
}

int main(){
	ios::sync_with_stdio(false);
	cin >> n >> q;
	rep(i, n-1) cin >> a[i + 1], a[i + 1]--;
	for(int i = 0; i < n; i += BS){
		rebuild(i / BS);
	}
	while(q--){
		int tp;
		cin >> tp;
		if(tp == 1){
			int l, r, x;
			cin >> l >> r >> x;
			l--, r--;
			for(int bn = l / BS; bn <= r / BS; bn++){
				if(bn * BS >= l && (bn + 1) * BS - 1 <= r && !has_to[bn]) offset[bn] = qmin(offset[bn] + x, n);
				else {
					for(int i = qmax(l, bn * BS); i < qmin((bn + 1) * BS, r + 1); i++) a[i] = qmax(0, a[i] - x);//, cout << "-" << i << endl;
					rebuild(bn);
				}
			}
		} else {
			int x, y;
			cin >> x >> y;
			x--, y--;
//			cout << "now: " << x << " " << y << endl;
			int lastx = x, lasty = y;
			while(x != y){
				if(x > y) lastx = x, go_block(x);
				else lasty = y, go_block(y);
			}
//			cout << "now: " << x << " " << y << endl;
			x = lastx, y = lasty;
			while(x != y){
				if(x > y) x = qmax(0, a[x] - offset[x / BS]);
				else y = qmax(0, a[y] - offset[y / BS]);
			}
//			cout << "now: " << x << " " << y << endl;
			cout << x + 1 << endl;
		}
	}
	return 0;
}