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 <algorithm>
using namespace std;
#define endl '\n'
#define ll long long
#define pi pair<int, int>
#define f first
#define s second

const int mxn = 100000, k = 400;
int n, q;
int a[mxn], b[mxn], f[mxn], p[mxn];

void upd(int x){
	b[x / k] += f[x] == x;
	f[x] = a[x / k] <= p[x] && (p[x] - a[x / k]) / k == x / k ? f[p[x]] : x;
	b[x / k] -= f[x] == x;
}

void add(int w, int x, int y, int z){
	for(int i = w * k + x; i <= w * k + y; i++) p[i] -= z;
	for(int i = w * k + x; i < min(n, (w + 1) * k); i++) upd(i);
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> q;
	
	for(int i = 1; i < n; i++) cin >> p[i], p[i]--, f[i] = i, upd(i);
	
	while(q--){
		int t;
		cin >> t;
		if(t & 1){
			int x, y, z;
			cin >> x >> y >> z;
			x--, y--;
			if(x / k == y / k) add(x / k, x % k, y % k, z);
			else add(x / k, x % k, k - 1, z), add(y / k, 0, y % k, z);
			for(int i = x / k + 1; i < y / k; i++){
				if(b[i]) add(i, 0, k - 1, z);
				else a[i] += z;
			}
		}else{
			int x, y;
			cin >> x >> y;
			x--, y--;
			while(x != y){
				if(x < y) swap(x, y);
				x = max(0, p[f[x] == f[y] ? x : f[x]] - a[x / k]);
			}
			cout << (x + 1) << endl;
		}
	}
	
	return 0;
}