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:
//In the name of Allah
#include <bits/stdc++.h>
using namespace std;

const int Sq = 321, N = Sq * Sq;
long long n, q, a[N], last[N], lazy[Sq], cnt[Sq];

int parent(int u) {
	return max(0ll, a[u] - lazy[u / Sq]);
}

void handle(int l, int val = 1) {
	cnt[l / Sq] += val;
	if (cnt[l / Sq] > Sq)
		return;
	for (int u = l; u < l + Sq && u < n; u++) 
		last[u] = parent(u) < l? u: last[parent(u)];
}

void update(int l, int r, int val) {
	int L = l, R = r;
	while (l < r)
		if (l % Sq != 0 || l + Sq > r) {
			a[l] -= val;
			l++;
		}
		else {
			lazy[l / Sq] += val;
			handle(l);
			l += Sq;
		}
	handle(L / Sq * Sq, 0);
	handle((R - 1) / Sq * Sq, 0);
}

int lca(int u, int v) {
	while (u != v) {
		u = parent(u / Sq != v / Sq || last[u] != last[v]? last[u]: u);
		if (u < v)
			swap(u, v);
	}
	return u;
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n >> q;
	for (int i = 1; i < n; i++) {
		cin >> a[i];
		a[i]--;
	}
	for (int i = 0; i < n; i += Sq)
		handle(i);
	while (q--) {
		int type;
		cin >> type;
		if (type == 1) {
			int l, r, x;
			cin >> l >> r >> x;
			update(--l, r, x);
		}
		else {
			int u, v;
			cin >> u >> v;
			if (u < v)
				swap(u, v);
			cout << lca(--u, --v) + 1 << endl;
		}
	}
}