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 pb push_back
#define fors(i, a, b) for(int i = (a); i < (b); ++i)
using namespace std;
const int maxn = 1e5 + 5;
const int B = 320;
int g[maxn], a[maxn], res[maxn], on[maxn], bel[maxn];
int in[maxn];
int n, q;
inline void go(int i, int x){
    if(g[i]==i){in[i]+=x; in[i] = min(in[i], n); return;}
    a[i] = max(a[i]-x, 1);
    if(bel[a[i]] == bel[i]) g[i] = g[a[i]];
    else g[i] = i, res[bel[i]]--;
    return;
}
int h[maxn];
void update(int l, int r, int x){
    int id1=bel[l], id2=bel[r];
    int ed = min((id2+1)*B-1, n);
    if(id1 == id2){
        while(l<=r) go(l++, x);
        fors(i,r+1,ed+1) go(i, 0);
        return;
    }
    while(l < (id1+1)*B) go(l++,x);
    fors(i,id2*B,r+1) go(i, x);
    fors(i,r+1,ed+1) go(i, 0);
    fors(i,id1+1,id2){
        if(res[i]==0) on[i]+=x, on[i] = min(on[i], n);
        else fors(j,i*B,(i+1)*B) go(j, x);
    }return;
}
inline int fa(int x){
    int t = in[x]+on[x/B];
    return max(a[x]-t, 1);
}
int lca(int x, int y){
    while(x!=y){
        if(x>y) swap(x,y);
        if(bel[x]!=bel[y]) y=fa(g[y]);
        else if(g[x]!=g[y]) x=fa(g[x]),y=fa(g[y]);
        else y=fa(y);
    }
    return x;
}
int main()
{
    scanf("%d%d", &n, &q); a[1] = 1; g[1] = 1;
    fors(i,1,n+1) bel[i] = i/B;
    fors(i,2,n+1) {
        scanf("%d", &a[i]);
        if(bel[a[i]] != bel[i]) g[i] = i;
        else g[i] = g[a[i]], res[bel[i]]++;
    }
    while(q--){
        int op,x,y; scanf("%d%d%d", &op,&x,&y);
        if(op==1){
            int w; scanf("%d", &w);
            update(x, y, w);
        }else{
            printf("%d\n", lca(x,y));
        }
    }
	return 0;
}