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 <iostream>
#include <vector>
#include <set>

using namespace std;

const int BLOCK_SIZE = 256;
const int MAX_N = 100007;


long long p[MAX_N], P[MAX_N]; //parent, specialParent
long long offset[MAX_N / BLOCK_SIZE + 5];
int n, q;
bool parentNearby[MAX_N];
set<int>pending[MAX_N / BLOCK_SIZE + 5];

int block(int x) {
    return x >> 8;
}

long long getp(int x) {
    return max((long long)0, p[x] - offset[block(x)]);
}

int getP(int x) {
    if(parentNearby[x])
        return P[x];
    return getp(x);
}

void Subtract(int a, int b, long long x) {
    int i = a;
    for(; i < n && block(i) == block(a); i++) {
        if(i <= b) {
            p[i] -= x;
            if(block(getp(i)) < block(i) && parentNearby[i]) {
                pending[block(i)].erase(pending[block(i)].find(i));
                parentNearby[i] = false;
            }
        }
        if(parentNearby[i]) {
            P[i] = getP(getp(i));
        }
    }
    for(; block(i) < block(b); i += BLOCK_SIZE) {
        offset[block(i)] += x;
        int blockId = block(i);
        vector<int>toErase;
        for(auto it = pending[blockId].begin(); it != pending[blockId].end(); it++) {
            if(block(getp(*it)) < blockId) {
                toErase.push_back(*it);
                parentNearby[*it] = false;
            } else {
                P[*it] = getP(getp(*it));
            }
        }
        
        for(int v : toErase) {
            pending[blockId].erase(pending[blockId].find(v));
        }
    }
    
    for(; i < n && block(i) == block(b); i++) {
        if(i <= b) {
            p[i] -= x;
            if(block(getp(i)) < block(i) && parentNearby[i]) {
                pending[block(i)].erase(pending[block(i)].find(i));
                parentNearby[i] = false;
            }
        }
        if(parentNearby[i]) {
            P[i] = getP(getp(i));
        }
    }
}

int LCA(int a, int b) {
    while(getP(a) != getP(b)) {
        if(a < b)
            swap(a, b);
        a = getP(a);
    }
    while(a != b) {
        if(a < b)
            swap(a, b);
        a = getp(a);
    }
    return a;
}


int main() {
    ios_base::sync_with_stdio(0);
    cin >> n >> q;
    p[0] = P[0] = 0;
    for(int i = 1; i < n; i++) {
        cin >> p[i];
        p[i]--;
        if(block(p[i]) != block(i) || block(i) == 0) {
            P[i] = getp(i);
        } else {
            P[i] = getP(getp(i));
            pending[block(i)].insert(i);
            parentNearby[i] = true;
        }
    }
    
    while(q--) {
        int t, a, b, x;
        cin >> t;
        if(t == 1) { //subtract
            cin >> a >> b >> x;
            Subtract(a - 1, b - 1, x);
        } else { //lca
            cin >> a >> b;
            cout << LCA(a - 1, b - 1) + 1 << '\n';
        }
    }
    
    return 0;
}

/*
 6 4
 1 2 3 3 4
 2 3 4
 1 2 3 1
 2 5 6
 2 2 3
*/