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 n, q, x;
int stsize, bstart;
long long segtree[300300 * 4];
void upd(int ind, long long v) {
  ind = bstart + ind;
  while (ind >= 0) {
    if (ind >= bstart)
      segtree[ind] = v;
    else
      segtree[ind] = segtree[ind * 2 + 1] + segtree[ind * 2 + 2];
    if (ind == 0) break;
    ind = (ind - 1) / 2;
  }
}
long long queryutil(int c, int cmin, int cmax, int minb, int maxb) {
  if (cmin >= minb && cmax <= maxb) return segtree[c];
  if (cmin > maxb || cmax < minb) return 0;
  return queryutil(c * 2 + 1, cmin, (cmin + cmax) / 2, minb, maxb) +
         queryutil(c * 2 + 2, 1 + (cmin + cmax) / 2, cmax, minb, maxb);
}
long long query(int c, int cmin, int cmax, int minb, int maxb) {
  int l = 1, k = 0;
  while (l < (cmax - cmin + 1)) l *= 2, k++;
  if (cmin >= minb && cmax <= maxb)
    return queryutil(0, 0, bstart, cmin ^ (x & (~((1 << k) - 1))),
                     cmax ^ (x & (~((1 << k) - 1))));
  if (cmin > maxb || cmax < minb) return 0;
  return query(c * 2 + 1, cmin, (cmin + cmax) / 2, minb, maxb) +
         query(c * 2 + 2, 1 + (cmin + cmax) / 2, cmax, minb, maxb);
}
int main() {
  scanf(" %d%d", &n, &q);
  x = 0;
  n = 1 << n;
  bstart = 1;
  while (bstart < n) bstart *= 2;
  stsize = bstart * 2 - 1;
  bstart--;
  memset(segtree, 0, sizeof(segtree));
  for (int i = 0; i < n; i++) {
    long long v;
    scanf("%lld", &v);
    upd(i, v);
  }
  for (int i = 0; i < q; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int ind;
      long long newv;
      scanf(" %d%lld", &ind, &newv);
      ind--;
      upd(ind ^ x, newv);
    } else if (type == 2) {
      int k;
      scanf(" %d", &k);
      x = x ^ ((1 << k) - 1);
    } else if (type == 3) {
      int k;
      scanf(" %d", &k);
      x = x ^ (1 << k);
    } else {
      int l, r;
      scanf(" %d%d", &l, &r);
      printf("%lld\n", query(0, 0, bstart, l - 1, r - 1));
    }
  }
}
