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 <cmath>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#define ll long long
using namespace std;
inline int read()
{
  int f = 1, x = 0;
  char ch;
  do
  {
    ch = getchar();
    if (ch == '-')
      f = -1;
  } while (ch < '0' || ch > '9');
  do
  {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * x;
}
const int N = 1e5;
const int M = 350;

int n, q, k;
int inchunk[N + 1], pa[N + 1], jumpchunk[N + 1];
ll tag[M + 1];

inline int par(int x) { return max(1ll, pa[x] - tag[inchunk[x]]); }
inline int st(int i) { return (i - 1) * k + 1; }
inline int ed(int i) { return min(i * k, n); }
inline void calc_jump(int l, int r)
{
  for (int i = l; i <= r; i++)
  {
    if (inchunk[i] == 1)
      jumpchunk[i] = 1;
    else
    {
      int p = par(i);
      if (inchunk[p] == inchunk[i])
        jumpchunk[i] = jumpchunk[p];
      else
        jumpchunk[i] = i;
    }
  }
  return;
}
inline void update(int l, int r, int x)
{
  if (inchunk[l] == inchunk[r])
  {
    for (int j = l; j <= r; j++)
      pa[j] = max(1, pa[j] - x);
    calc_jump(st(inchunk[l]), ed(inchunk[l]));
  }
  else
  {
    for (int j = l; j <= ed(inchunk[l]); j++)
      pa[j] = max(1, pa[j] - x);
    calc_jump(st(inchunk[l]), ed(inchunk[l]));
    for (int j = st(inchunk[r]); j <= r; j++)
      pa[j] = max(1, pa[j] - x);
    calc_jump(st(inchunk[r]), ed(inchunk[r]));
    for (int j = inchunk[l] + 1; j < inchunk[r]; j++)
    {
      tag[j] += x;
      if (tag[j] - x <= k)
        calc_jump(st(j), ed(j));
    }
  }
}
inline void LCA(int l, int r)
{
  while (jumpchunk[l] != jumpchunk[r])
  {
    if (jumpchunk[l] > jumpchunk[r])
      swap(l, r);
    r = par(jumpchunk[r]);
  }
  while (l != r)
  {
    if (l > r)
      swap(l, r);
    r = par(r);
  }
  printf("%d\n", l);
}
inline void query(int q)
{
  for (int i = 1; i <= q; i++)
  {
    int opt = read(), l = read(), r = read();
    if (opt == 1)
    {
      int x = read();
      update(l, r, x);
    }
    else
      LCA(l, r);
  }
}
int main()
{
  n = read(), q = read();
  for (int i = 2; i <= n; i++)
    pa[i] = read();
  k = sqrt(n);
  for (int i = 1; i <= n; i++)
    inchunk[i] = (i - 1) / k + 1;
  calc_jump(1, n);

  query(q);
  return 0;
}