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 N = 1e5 + 3;
const int C = 300;

int n, q, a[N], lp[N], fa[N];
bool b[N];

int get_bucket(int i){
    return (i - 1) / C;
}

void update_all(int i, int x){
    if(b[i]){
        lp[i] += x;
        return;
    }

    b[i] = true;
    for(int j = i * C + 1; j <= (i + 1) * C; ++j){
        a[j] = max(1, a[j] - x);
        fa[j] = j;
        if(a[j] < i * C + 1) fa[j] = j;
        else fa[j] = fa[a[j]];

        if(fa[j] != j) b[i] = false;
    }
}

void update_part(int i, int l, int r, int x){
    if(b[i]){
        for(int j = l; j <= r; ++j)
            a[j] = max(1, a[j] - x);
        return;
    }

    b[i] = true;
    for(int j = i * C + 1; j <= min((i + 1) * C, n); ++j){
        if(l <= j && j <= r) a[j] = max(1, a[j] - x);
        fa[j] = j;
        if(a[j] < i * C + 1) fa[j] = j;
        else fa[j] = fa[a[j]];

        if(fa[j] != j) b[i] = false;
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> q;
    for(int i = 2; i <= n; ++i)
        cin >> a[i];

    a[1] = 1;
    
    for(int i = 0; i <= get_bucket(n); ++i){
        int idx = i * C + 1;
        fa[idx] = idx;

        b[i] = true;
        for(int j = idx + 1; j <= min(n, (i + 1) * C); ++j){
            fa[j] = j;
            if(a[j] < idx) fa[j] = j;
            else fa[j] = fa[a[j]];

            if(fa[j] != j) b[i] = false;
        }
    }

    for(int i = 1; i <= q; ++i){
        int t;
        cin >> t;

        if(t == 1){
            int l, r, x;
            cin >> l >> r >> x;
            int bl = get_bucket(l), br = get_bucket(r);
            for(int i = bl + 1; i < br; ++i)
                update_all(i, x);

            if(bl == br) update_part(bl, l, r, x);
            else{
                update_part(bl, l, (bl + 1) * C, x);
                update_part(br, br * C + 1, r, x);
            }
        }
        else{
            int u, v;
            cin >> u >> v;
            while(u != v){
                if(u < v) swap(u, v);
                if(fa[u] >= v){
                    u = fa[u];
                    if(u == v) break;
                    u = max(1, a[u] - lp[get_bucket(u)]);
                    continue;
                }
                if(fa[u] != fa[v]){
                    u = a[fa[u]];
                    v = a[fa[v]];
                    continue;
                }
                while(u != v){
                    u = a[u];
                    if(u < v) swap(u, v);
                }
            }

            cout << u << "\n";
        }
    }
}