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>

using namespace std;
const int c=105002, gy=320;
int n, q, t[c], fel[c], lp[c];
int mbl[c], kezd[c], veg[c], jo[c];
void valt(int a, int b) {
    t[a]=max(1, t[a]-b);
}
int kov(int a) {
    return max(1, t[a]-lp[mbl[a]]);
}
void solve(int bl, int ert) {
    if (jo[bl]) {
        lp[bl]+=ert;
        lp[bl]=min(lp[bl], n);
        return;
    }
    bool sp=1;
    for (int i=kezd[bl]; i<=veg[bl]; i++) {
        valt(i, ert);
        int s=kov(i);
        if (mbl[s]!=mbl[i]) {
            fel[i]=i;
        } else {
            fel[i]=fel[s];
            sp=0;
        }
    }
    if (sp) {
        jo[bl]=1;
    }
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin >> n >> q;
    for (int i=1; i<=n; i++) {
        mbl[i]=mbl[i-1]+(((i-1)%gy)==0);
    }
    for (int i=1; i<=n; i++) {
        kezd[i]=(i-1)*gy+1, veg[i]=i*gy;
    }
    for (int i=2; i<=n; i++) {
        cin >> t[i];
    }

    for (int i=1; kezd[i]<=n; i++) {
        solve(i, 0);
    }
    for (int i=1; i<=q; i++) {
        int s;
        cin >> s;
        if (s==1) {
            int l, r, x;
            cin >> l >> r >> x;
            int kb=mbl[l], nb=mbl[r];
            if (kb==nb) {
                for (int i=l; i<=r; i++) {
                    valt(i, x);
                }
                solve(kb, 0);
            } else {
                while (mbl[l]==kb) {
                    valt(l, x);
                    l++;
                }
                solve(kb, 0);
                while (mbl[r]==nb) {
                    valt(r, x);
                    r--;
                }
                solve(nb, 0);
                for (int i=kb+1; i<nb; i++) {
                    solve(i, x);
                }
            }
        }
        if (s==2) {
            int a, b;
            cin >> a >> b;
            if (a<b) {
                swap(a, b);
            }
            while (fel[a]!=fel[b]) {
                a=kov(fel[a]);
                if (a<b) {
                    swap(a, b);
                }
            }
            while (a!=b) {
                a=kov(a);
                if (a<b) {
                    swap(a, b);
                }
            }
            cout << a << "\n";
        }
    }
    return 0;
}
/*
6 4
1 2 3 3 4
2 3 4
1 2 3 1
2 5 6
2 2 3
*/
