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;
const int N = 200005;
long long t[(1 << 18) * 4 + 10], rev[20];
int n, k, q;
void add(int d, int v, int l, int r, int pos, int val) {
  if (l == r) {
    t[v] = val;
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m)
    add(d + 1, (v + v) ^ rev[d], l, m, pos, val);
  else
    add(d + 1, (v + v + 1) ^ rev[d], m + 1, r, pos, val);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
long long get(int d, int v, int l, int r, int i, int j) {
  if (i > j) return 0;
  if (l == i && r == j) return t[v];
  int m = (l + r) / 2;
  return get(d + 1, (v + v) ^ rev[d], l, m, i, min(m, j)) +
         get(d + 1, (v + v + 1) ^ rev[d], m + 1, r, max(i, m + 1), j);
}
int main() {
  cin >> k >> q;
  n = (1 << k);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    add(0, 1, 0, n - 1, i, a);
  }
  while (q--) {
    int T;
    cin >> T;
    if (T == 1) {
      int x, y;
      cin >> x >> y;
      x--;
      add(0, 1, 0, n - 1, x, y);
    } else if (T == 2) {
      int x;
      cin >> x;
      x = k - x;
      for (int i = x; i <= k; i++) rev[i] ^= 1;
    } else if (T == 3) {
      int x;
      cin >> x;
      x = k - x - 1;
      rev[x] ^= 1;
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << get(0, 1, 0, n - 1, l, r) << "\n";
    }
  }
  return 0;
}
