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;
int64_t tree[(1 << 20) + 1] = {-1};
int size;
void buildTree(int64_t *arr, int left, int right, int root) {
  if (left == right)
    tree[root] = arr[left];
  else {
    int mid = (left + right) / 2;
    buildTree(arr, left, mid, root * 2 + 1);
    buildTree(arr, mid + 1, right, root * 2 + 2);
    tree[root] = tree[root * 2 + 1] + tree[root * 2 + 2];
  }
}
int64_t replace(int64_t val, int pos, int left, int right, int root) {
  if (left == right && pos == left) {
    int64_t delta = val - tree[root];
    tree[root] = val;
    return delta;
  } else {
    int mid = (left + right) / 2;
    int64_t delta;
    if (pos > mid)
      delta = replace(val, pos, mid + 1, right, root * 2 + 2);
    else
      delta = replace(val, pos, left, mid, root * 2 + 1);
    tree[root] += delta;
    return delta;
  }
}
int64_t sum(int ql, int qr, int left, int right, int root, int x) {
  if (qr < left || ql > right) {
    return 0;
  }
  if (ql <= left && qr >= right) {
    int len = right - left + 1;
    int delta = (x ^ left) / len - (left / len);
    return tree[root + delta];
  } else {
    int mid = (left + right) / 2;
    return sum(ql, qr, left, mid, root * 2 + 1, x) +
           sum(ql, qr, mid + 1, right, root * 2 + 2, x);
  }
}
int main(void) {
  int n, q;
  cin >> n >> q;
  int size = 1 << n;
  int64_t *arr = new int64_t[1 << n];
  for (int i = 0; i < size; i++) {
    cin >> arr[i];
  }
  buildTree(arr, 0, size - 1, 0);
  int op;
  int x = 0;
  int val, pos;
  while (cin >> op) {
    switch (op) {
      case 1:
        cin >> pos >> val;
        pos--;
        replace(val, pos ^ x, 0, size - 1, 0);
        break;
      case 2:
        cin >> val;
        x ^= ((1 << val) - 1);
        break;
      case 3:
        cin >> val;
        x ^= (1 << val);
        break;
      case 4:
        int ql, qr;
        cin >> ql >> qr;
        ql--;
        qr--;
        cout << sum(ql, qr, 0, size - 1, 0, x) << endl;
        break;
    }
  }
  return 0;
}
