You are given an array a of length 2^n. You should process q queries on it. Each query has one of the following 4 types: 

  1. Replace(x, k) — change a_x to k; 
  2. Reverse(k) — reverse each subarray [(i-1) ⋅ 2^k+1, i ⋅ 2^k] for all i (i ≥ 1); 
  3. Swap(k) — swap subarrays [(2i-2) ⋅ 2^k+1, (2i-1) ⋅ 2^k] and [(2i-1) ⋅ 2^k+1, 2i ⋅ 2^k] for all i (i ≥ 1); 
  4. Sum(l, r) — print the sum of the elements of subarray [l, r]. 



Write a program that can quickly process given queries.

Input

The first line contains two integers n, q (0 ≤ n ≤ 18; 1 ≤ q ≤ 10^5) — the length of array a and the number of queries.

The second line contains 2^n integers a_1, a_2, …, a_{2^n} (0 ≤ a_i ≤ 10^9).

Next q lines contains queries — one per line. Each query has one of 4 types: 

  * "1 x k" (1 ≤ x ≤ 2^n; 0 ≤ k ≤ 10^9) — Replace(x, k); 
  * "2 k" (0 ≤ k ≤ n) — Reverse(k); 
  * "3 k" (0 ≤ k < n) — Swap(k); 
  * "4 l r" (1 ≤ l ≤ r ≤ 2^n) — Sum(l, r). 



It is guaranteed that there is at least one Sum query.

Output

Print the answer for each Sum query.

Examples

Input


2 3
7 4 9 9
1 2 8
3 1
4 2 4


Output


24


Input


3 8
7 0 8 8 7 1 5 2
4 3 7
2 1
3 2
4 1 6
2 3
1 5 16
4 8 8
3 0


Output


29
22
1

Note

In the first sample, initially, the array a is equal to \{7,4,9,9\}.

After processing the first query. the array a becomes \{7,8,9,9\}.

After processing the second query, the array a_i becomes \{9,9,7,8\}

Therefore, the answer to the third query is 9+7+8=24.

In the second sample, initially, the array a is equal to \{7,0,8,8,7,1,5,2\}. What happens next is: 

  1. Sum(3, 7) → 8 + 8 + 7 + 1 + 5 = 29; 
  2. Reverse(1) → \{0,7,8,8,1,7,2,5\}; 
  3. Swap(2) → \{1,7,2,5,0,7,8,8\}; 
  4. Sum(1, 6) → 1 + 7 + 2 + 5 + 0 + 7 = 22; 
  5. Reverse(3) → \{8,8,7,0,5,2,7,1\}; 
  6. Replace(5, 16) → \{8,8,7,0,16,2,7,1\}; 
  7. Sum(8, 8) → 1; 
  8. Swap(0) → \{8,8,0,7,2,16,1,7\}. 
#include <bits/stdc++.h>
using namespace std;
int arr[270000];
int rev[20];
void reverse(int k) {
  for (int i = k; i >= 0; i--) rev[i] ^= 1;
}
struct node {
  node *lch, *rch;
  long long sum;
  int tag;
  node(int val = 0) {
    lch = rch = nullptr;
    sum = val;
    tag = 0;
  }
  void build(int l, int r) {
    if (l == r) {
      sum = arr[l];
    } else {
      int mid = (l + r) >> 1;
      lch = new node();
      lch->build(l, mid);
      rch = new node();
      rch->build(mid + 1, r);
      sum = lch->sum + rch->sum;
    }
  }
  void push(int k) {
    if (rev[k] ^ tag == 1) {
      swap(lch, rch);
      tag ^= 1;
    }
  }
  void modify(int l, int r, int pos, int new_val, int k) {
    if (l == r)
      sum = new_val;
    else {
      int mid = (l + r) >> 1;
      push(k);
      if (pos <= mid)
        lch->modify(l, mid, pos, new_val, k - 1);
      else
        rch->modify(mid + 1, r, pos, new_val, k - 1);
      sum = lch->sum + rch->sum;
    }
  }
  long long query(int l, int r, int ql, int qr, int k) {
    if (l == ql && r == qr)
      return sum;
    else {
      int mid = (l + r) >> 1;
      push(k);
      if (qr <= mid)
        return lch->query(l, mid, ql, qr, k - 1);
      else if (ql > mid)
        return rch->query(mid + 1, r, ql, qr, k - 1);
      else
        return lch->query(l, mid, ql, mid, k - 1) +
               rch->query(mid + 1, r, mid + 1, qr, k - 1);
    }
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  int m = (1 << n);
  for (int i = 1; i <= m; i++) cin >> arr[i];
  node *root = new node();
  root->build(1, m);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int x, k;
      cin >> x >> k;
      root->modify(1, m, x, k, n);
    } else if (type == 2) {
      int k;
      cin >> k;
      reverse(k);
    } else if (type == 3) {
      int k;
      cin >> k;
      reverse(k + 1);
      reverse(k);
    } else {
      int l, r;
      cin >> l >> r;
      cout << root->query(1, m, l, r, n) << '\n';
    }
  }
  return 0;
}
