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>

using namespace std;

const int MAXN = 100005;
const int MAXM = 321;

int n, siz, q, a[MAXN], b[MAXN];
int lft[MAXM], rit[MAXM], bel[MAXN], tag[MAXM], cnt[MAXM];

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
	while (isdigit(ch)) { (x *= 10) += ch - '0'; ch = getchar(); }
	return x * f;
}

inline void update(int x)
{
	for (int i = lft[x]; i <= rit[x]; i++)
		a[i] = max(1, a[i]-tag[x]); tag[x] = 0;
	for (int i = lft[x]; i <= rit[x]; i++)
	{
		if (a[i] < lft[x]) b[i] = a[i];
		else b[i] = b[a[i]];
	}
}

inline void prepare()
{
	siz = sqrt(n);
	for (int i = 1; i <= (n-1)/siz+1; i++)
	{
		lft[i] = rit[i-1]+1, rit[i] = min(rit[i-1]+siz, n);
		for (int j = lft[i]; j <= rit[i]; j++)
			bel[j] = i;
	}
	for (int i = 1; i <= bel[n]; i++)
		update(i);
}

void write(int x)
{
	if (x > 9) write(x / 10);
	putchar('0' + x % 10);
}

inline void upd(int l, int r, int x)
{
	if (bel[l] == bel[r])
	{
		for (int i = l; i <= r; i++)
			a[i] = max(1, a[i]-x);
		update(bel[l]); return;
	}
	for (int i = l; i <= rit[bel[l]]; i++)
		a[i] = max(1, a[i]-x); update(bel[l]);
	for (int i = lft[bel[r]]; i <= r; i++)
		a[i] = max(1, a[i]-x); update(bel[r]);
	for (int i = bel[l]+1; i <= bel[r]-1; i++)
	{
		tag[i] = min(n, tag[i]+x); cnt[i]++;
		if (cnt[i] <= siz+2) update(i);
	}
}

inline int get_b(int x)
{ return max(1, b[x]-tag[bel[x]]); }

inline int query(int x, int y)
{
	while (get_b(x) != get_b(y) || bel[x] != bel[y])
	{
		if (bel[x] != bel[y])
		{
			if (bel[x] < bel[y]) swap(x, y);
			x = get_b(x);	
		}
		else
			x = get_b(x), y = get_b(y);
	}
	while (x != y)
	{
		if (x < y) swap(x, y);
		x = max(a[x]-tag[bel[x]], 1);
	}
	return x;
}

int main()
{
	n = read(), q = read();
	for (int i = 2; i <= n; i++)
		a[i] = read();
		
	prepare();
	
	for (int i = 1, opt, x, y, z; i <= q; i++)
	{
		opt = read();
		if (opt == 1)
			x = read(), y = read(), z = read(), upd(x, y, z);
		if (opt == 2)
			x = read(), y = read(), write(query(x, y)), putchar('\n');
	}

  return 0;
}