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 <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define eps 1e-8
#define maxm 600005 
#define maxn 100005
#define ls (tot << 1)
#define rs (tot << 1 | 1)
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const double pi = acos(-1);
const ll mod = 1e9 + 7;
inline ll read(){
    ll x = 0, f = 1;char ch = getchar();
    while(ch > '9' || ch < '0'){if(ch == '-') f = -1;ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = x * 10 + ch -'0';ch = getchar();}
    return x * f;
}
int n, q, a[maxn];
int B = 300, cnt = 0, f[maxn], l[maxn], r[maxn], bl[maxn], lazy[maxn];
bool flag[maxn];
void work(int id){
	flag[id] = 0;
	for(int i = l[id];i <= r[id];i++){
		if(a[i] < l[id]) f[i] = i;
		else f[i] = f[a[i]], flag[id] = 1;
	}
}
void work(int L, int R, int x){
	for(int i = L;i <= R;i++)
		a[i] = i == 1 ? 0 : max(1, a[i] - x);
	work(bl[L]);
}
void work(int id, int x){
	if(!flag[id]) lazy[id] += x;
	else work(l[id], r[id], x);
}
int query(int L, int R){
	while(f[L] != f[R]){
		if(L < R) swap(L, R);
		L = max(1, a[f[L]] - lazy[(L - 1) / B + 1]);
	}
	while(L != R){
		if(L < R) swap(L, R);
		L = max(1, a[L] - lazy[(L - 1) / B + 1]);
	}
	return L;
}
int main(){
	int i, j, op, L, R, x;
	n = read(), q = read();
	bl[1] = 1;
	for(i = 2;i <= n;i++){
		a[i] = read();
		bl[i] = (i - 1) / B + 1;
	}
	cnt = (n - 1) / B + 1;
	for(i = 1;i <= cnt;i++){
		l[i] = (i - 1) * B + 1, r[i] = min(n, i * B);
		work(i);
	}
	while(q--){
		op = read(), L = read(), R = read();
		if(op == 1){
			x = read();
			if(bl[L] == bl[R]) work(L, R, x);
			else{
				work(L, r[bl[L]], x);
				for(i = bl[L] + 1;i < bl[R];i++){
					work(i, x);
				}
				work(l[bl[R]], R, x);
			}
		}
		else{
			printf("%d\n", query(L, R));
		}
	}
	return 0;
}
/*
8
1 2
2 3
2 4
4 5
4 6
6 7
7 8
*/