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;

using ll = long long;

const int B = 100;

const int maxn = 100100;

int p[maxn];
int a[maxn], lazy[maxn / B + 5];

int main(void) {
    //freopen("h.in", "r", stdin);

    int n, q; scanf("%d%d", &n, &q);
    for (int i=1; i<n; i++) scanf("%d", &a[i]), a[i]--;
    for (int i=1; i<n; i++) p[i] = (i/B == a[i]/B) ? p[a[i]] : i;

    auto get_parent = [](int i) -> int { return max(a[i] - lazy[i/B], 0); };

    auto get_p = [&](int i) {
        for (int j=i*B, k=i*B+B; j<k; j++) {
            int pa = max(a[j] - lazy[i], 0);
            if (i == pa/B) p[j] = p[pa];
            else p[j] = j;
        }
    };

    for (int _q=0; _q<q; _q++) {
        int t, u, v, x; scanf("%d%d%d", &t, &u, &v);
        u--, v--;
        if (t == 1) {
            scanf("%d", &x);
            for (int i=u/B; i<=v/B; i++) {
                if (u <= i*B && i*B+B-1 <= v) {
                    if (lazy[i] < B) {
                        lazy[i] += x; if (lazy[i] > n) lazy[i] = n;
                        get_p(i);
                    } else {
                        lazy[i] += x; if (lazy[i] > n) lazy[i] = n;
                    }
                } else {
                    for (int j=max(u, i*B), j_end=min(v, i*B+B-1); j<=j_end; j++) {
                        a[j] -= x; if (a[j] < 0) a[j] = 0;
                    }
                    get_p(i);
                }
            }
        } else {
            while (u/B != v/B) {
                if (u < v) v = get_parent(p[v]);
                else u = get_parent(p[u]);
            }
            while (u != v) {
                if (u < v) v = get_parent(v);
                else u = get_parent(u);
            }
            printf("%d\n", u + 1);
        }
    }
    return 0;
}