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 cmp(const void* a, const void* b) { return *(int*)a - *(int*)b; }
const int NUM = 1 << 18;
long long a[19][NUM];
int len[19], swaps[19], n, num;
void init() {
  for (int i = 0; i < n; i++) {
    len[i] = 1 << (n - i);
    for (int j = 0; j < len[i]; j++) {
      a[i + 1][j >> 1] += a[i][j];
    }
  }
  len[n] = 1;
}
void swap(int level) {
  level++;
  swaps[level] = 1 - swaps[level];
}
void reverse(int level) {
  while (level > 0) {
    swaps[level] = 1 - swaps[level];
    level--;
  }
}
int getInitialPos(int pos, int level, int index) {
  if (level == 0) return index;
  int half = 1 << (level - 1);
  if (swaps[level] == 0) {
    if (pos < half)
      return getInitialPos(pos, level - 1, index << 1);
    else
      return getInitialPos(pos - half, level - 1, (index << 1) + 1);
  } else {
    if (pos >= half)
      return getInitialPos(pos - half, level - 1, index << 1);
    else
      return getInitialPos(pos, level - 1, (index << 1) + 1);
  }
}
void replace(int pos, int val) {
  int old = pos;
  pos = getInitialPos(pos, n, 0);
  int delta = val - a[0][pos];
  for (int i = 0; i <= n; i++) {
    a[i][pos] += delta;
    pos >>= 1;
  }
}
long long query(int pos, int level, int index) {
  if (pos < 0) return 0;
  if (level == 0) return a[0][index];
  int half = 1 << (level - 1);
  long long res = 0;
  if (swaps[level] == 0) {
    if (pos < half)
      res = query(pos, level - 1, index << 1);
    else
      res = a[level - 1][index << 1] +
            query(pos - half, level - 1, (index << 1) + 1);
  } else {
    if (pos >= half)
      res = a[level - 1][(index << 1) + 1] +
            query(pos - half, level - 1, index << 1);
    else
      res = query(pos, level - 1, (index << 1) + 1);
  }
  return res;
}
int main() {
  int q, op, x, k, l, r;
  long long res = 0, res1, res2;
  scanf("%d%d", &n, &q);
  num = 1 << n;
  for (int i = 0; i < num; i++) scanf("%lld", &a[0][i]);
  init();
  while (q-- > 0) {
    scanf("%d", &op);
    switch (op) {
      case 1:
        scanf("%d%d", &x, &k);
        replace(x - 1, k);
        break;
      case 2:
        scanf("%d", &k);
        reverse(k);
        break;
      case 3:
        scanf("%d", &k);
        swap(k);
        break;
      case 4:
        scanf("%d%d", &l, &r);
        res1 = query(r - 1, n, 0);
        res2 = query(l - 2, n, 0);
        res = res1 - res2;
        printf("%lld\r\n", res);
        break;
      default:
        break;
    }
  }
  return 0;
}
