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 c = (1 << 19);
int n, q, ke[c], ve[c], xo;
long long sum[c];
long long solve(int a, int st, int fi) {
  int s = ke[a], f = ve[a];
  if (st > f || fi < s) return 0;
  if (st <= s && f <= fi) {
    int tav = f + 1 - s;
    return sum[a ^ (xo / tav)];
  }
  return solve(2 * a, st, fi) + solve(2 * a + 1, st, fi);
}
int main() {
  cin >> n >> q;
  n = (1 << n);
  for (int i = n; i < 2 * n; i++) cin >> sum[i], ke[i] = i - n, ve[i] = i - n;
  for (int i = n - 1; i >= 1; i--)
    sum[i] = sum[2 * i] + sum[2 * i + 1], ke[i] = ke[2 * i],
    ve[i] = ve[2 * i + 1];
  for (int i = 1; i <= q; i++) {
    int a, b, c;
    cin >> a >> b;
    if (a == 1) {
      cin >> c;
      b = n + xo ^ (b - 1), sum[b] = c;
      while (b > 1) {
        b /= 2;
        sum[b] = sum[2 * b] + sum[2 * b + 1];
      }
    }
    if (a == 2) xo = xo ^ ((1 << b) - 1);
    if (a == 3) xo = xo ^ (1 << b);
    if (a == 4) {
      cin >> c;
      b--, c--;
      cout.flush() << solve(1, b, c) << "\n";
    }
  }
  return 0;
}
