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 endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < int(n); i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb push_back

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 100005, SQ = 316;
int n, q, blo[MX];
ll lz[MX], act[MX], a[MX], to[MX];

void update (int i, int u, int v, int x) {
	a[i] -= lz[blo[i]] - act[i];
	act[i] = lz[blo[i]];
	
	if (u <= i && i <= v)
		a[i] -= x;
	
	if (a[i] < 1)
		a[i] = 1;
	
	if (blo[a[i]] == blo[i])
		to[i] = to[a[i]];
	else
		to[i] = a[i];
}

void update (int i) {
	a[i] -= lz[blo[i]] - act[i];
	act[i] = lz[blo[i]];
	
	if (a[i] < 1)
		a[i] = 1;
	
	if (blo[a[i]] == blo[i])
		to[i] = to[a[i]];
	else
		to[i] = a[i];
}

void refresh (int b) {
	for (int i = b * SQ; i <= n && blo[i] == b; i++)
		update(i);
}

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> q;
	
	a[1] = to[1] = 1;
	for (int i = 2; i <= n; i++) {
		cin >> a[i];
		blo[i] = i / SQ;
	}
	
	for (int i = 1; i <= n; i++)
		update(i);
		
	while (q--) {
		int t, u, v;
		cin >> t >> u >> v;
		
		if (t == 1) {
			int x;
			cin >> x;
			
			for (int i = SQ * blo[u]; i <= n && blo[i] == blo[u]; i++)
				update(i, u, v, x);
			
			for (int j = blo[u] + 1; j < blo[v]; j++)
				lz[j] += x;
			
			if (blo[u] != blo[v])
				for (int i = SQ * blo[v]; i <= n && blo[i] == blo[v]; i++)
					update(i, u, v, x);
		} else {
			while (u != v) {
				if (act[u] < lz[blo[u]]) refresh(blo[u]);
				if (act[v] < lz[blo[v]]) refresh(blo[v]);
				
				if (to[u] == to[v]) break;
				
				if (u == 1 || v == 1) {
					u = v = 1;
					break;
				}
				
				if (to[u] < to[v]) swap(u, v);
				u = to[u];
			}
			
			while (u != v) {
				if (act[u] < lz[blo[u]]) refresh(blo[u]);
				if (act[v] < lz[blo[v]]) refresh(blo[v]);
				
				if (u == 1 || v == 1) {
					u = v = 1;
					break;
				}
				
				if (a[u] < a[v]) swap(u, v);
				u = a[u];
			}
			
			cout << u << endl;
		}
	}
	
	return 0;
}