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 ll long long
#define db long double
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define all(a) a.begin(), a.end()

using namespace std;

const int MAGIC = 300;
const int BLOCKS = 160000/MAGIC;

vector<int> tree;
vector<int> leave_block;

int add[BLOCKS];
int cnt[BLOCKS];

int n;

void recalc(int block) {
  for (int i = block * MAGIC; i < min(n, (block + 1) * MAGIC); ++i) {
    tree[i] = max(0, tree[i] - add[block]);
  }
  add[block] = 0;

  for (int i = block * MAGIC; i < min(n, (block + 1) * MAGIC); ++i) {
    int W = tree[i];
    if (W < block * MAGIC) leave_block[i] = W;
    else leave_block[i] = leave_block[W];
  }

}

int layer(int x) {
  return x/MAGIC;
}

int go_outer(int x) {
  return max(0, leave_block[x] - add[layer(x)]);
}

int go(int x) {
  return max(0, tree[x] - add[layer(x)]);
}

int main(){
#ifdef LOCAL
	freopen("H_input.txt", "r", stdin);
	//freopen("H_output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(0);
	cin.tie(0);

  int q;
  cin >> n >> q;

  tree.assign(n, 0);
  for (int i = 1; i < n; ++i) {
    cin >> tree[i];
    tree[i]--;
  }

  leave_block.assign(n, 0);

  for (int i = 0; i * MAGIC < n; ++i) {
    recalc(i);
  }

  for (int i = 0; i < q; ++i) {
    int tp;
    cin >> tp;

    //cout << " === " << go_outer(4) << endl;

    if (tp == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      l--, r--;
      for (int j = 0; j * MAGIC < n; ++j) {
        int L = j*MAGIC, R = min(n, (j+1)*MAGIC) - 1;
        
        int pl = max(L, l), pr = min(R, r);
        if (pl > pr) continue;
        if (pl != L || pr != R || cnt[j] <= MAGIC) {
          if (pl == L && pr == R) {
            cnt[j]++;
            add[j]+=x;
            add[j] = min(add[j], (int) 1e8);
          }
          else {
            for (int e = pl; e <= pr; ++e) tree[e] = max(0, tree[e] - x);
          }
        //cout << " ??? " << j << endl;
          recalc(j);
          //cout << leave_block[3] << " " << tree[3] << " " << pl <<  " " << pr  << endl;
        } 
        else {
          add[j]+=x;
          add[j] = min(add[j], (int) 1e8);
        }

      }
    }

    else {
      int u, v;
      cin >> u >> v;

      u--, v--;
      while (true) {
        if (u == v) break;
        //cout << u << " " << v << endl;
        //cout << go_outer(u) << " " << go_outer(v) << endl;
        if (layer(u) < layer(v)) {
          swap(u, v);
        }
        if (layer(u) > layer(v)) {
          u = go_outer(u);
        }
        else {
          int A = go_outer(u), B = go_outer(v);
          if (A!=B) {
            u = A, v = B;
          }
          else {
            while (u != v) {
              if (u > v) u = go(u);
              else v = go(v); 
            }
          }
        }
      }

      cout << u+1 << '\n';


    }
  }


}