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 <cstdlib>
#include <cstdio>
#include <cmath>
#define ll long long
using namespace std;
inline int read()
{
	int f = 1, x = 0;
	char ch;

	do{
		ch = getchar();
		if (ch == '-')
			f = -1;
	}while(ch < '0' || ch > '9');
	do{
		x = x * 10 + ch - '0';
		ch = getchar();
	}while(ch >= '0' && ch <= '9');
	return f * x;
}
const int N = 1e5;
const int M = 350;

int n, q, k;
int ink[N + 1], st[M + 1], ed[M + 1], fa[N + 1], outk[N + 1];
ll tag[M + 1];

inline int gfa(int x)
{
	return max(1ll, fa[x] - tag[ink[x]]);
}
inline void calc(int l, int r)
{
	for (int i = l; i <= r; i++) {
		if (ink[i] == 1) {
			outk[i] = 1;
		} else {
			int f = gfa(i);
			if (ink[f] == ink[i])
				outk[i] = outk[f];
			else
				outk[i] = i;
		}
	}
	return;
}
int main()
{
	n = read();
	q = read();
	for (int i = 2; i <= n; i++)
		fa[i] = read();
	k = sqrt(n);
	for (int i = 1; i <= n; i++)
		ink[i] = (i - 1) / k + 1;
	for (int i = 1; i <= ink[n]; i++) {
		st[i] = (i - 1) * k + 1;
		ed[i] = min(i * k, n);
	}
	calc(1, n);

	for (int i = 1; i <= q; i++) {
		int opt = read(), l = read(), r = read();
		if (opt == 1) {
			int x = read();
			if (ink[l] == ink[r]) {
				for (int j = l; j <= r; j++)
					fa[j] = max(1, fa[j] - x);
				calc(st[ink[l]], ed[ink[l]]);
			} else {
				for (int j = l; j <= ed[ink[l]]; j++)
					fa[j] = max(1, fa[j] - x);
				calc(st[ink[l]], ed[ink[l]]);
				for (int j = st[ink[r]]; j <= r; j++)
					fa[j] = max(1, fa[j] - x);
				calc(st[ink[r]], ed[ink[r]]);
				for (int j = ink[l] + 1; j < ink[r]; j++) {
					tag[j] += x;
					if (tag[j] - x <= k)
						calc(st[j], ed[j]);
				}
			}
		} else {
			while (outk[l] != outk[r]) {
				if (outk[l] > outk[r])
					swap(l, r);
				r = gfa(outk[r]);
			}
			if (l == r) {
				printf("%d\n", l);
				continue;
			}
			while (gfa(l) != gfa(r)) {
				if (gfa(l) > gfa(r))
					swap(l, r);
				r = gfa(r);
			}
			if (l == r)
				printf("%d\n", l);
			else
				printf("%d\n", gfa(l));
		}
	}

	return 0;
}
