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 s;
long long seg[524288];
bool rev[20];
void upd(int i, int d, int x, int k) {
  if (d == -1) {
    seg[i] = k;
    return;
  }
  bool rgt = (x > (1 << d));
  if (rgt) x -= (1 << d);
  rgt ^= rev[d];
  upd(i * 2 + rgt, d - 1, x, k);
  seg[i] = seg[i * 2] + seg[i * 2 + 1];
}
long long query(int l, int r, int i, int d, int ql, int qr) {
  if (ql > r || qr < l)
    return 0;
  else if (ql <= l && r <= qr)
    return seg[i];
  int m = (l + r) >> 1;
  return query(l, m, i * 2 + rev[d], d - 1, ql, qr) +
         query(m + 1, r, i * 2 + !rev[d], d - 1, ql, qr);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  s = (1 << n);
  memset(seg, 0, sizeof(seg));
  for (int i = (1); i <= (s); i++) {
    scanf("%lld", &seg[s + i - 1]);
  }
  for (int i = (s - 1); i >= (1); i--) seg[i] = seg[i * 2] + seg[i * 2 + 1];
  for (int i = (1); i <= (q); i++) {
    int t, x, k, l, r;
    scanf("%d", &t);
    switch (t) {
      case 1:
        scanf("%d %d", &x, &k);
        upd(1, n - 1, x, k);
        break;
      case 2:
        scanf("%d", &k);
        for (int j = (k - 1); j >= (0); j--) rev[j] ^= 1;
        break;
      case 3:
        scanf("%d", &k);
        rev[k] ^= 1;
        break;
      case 4:
        scanf("%d %d", &l, &r);
        printf("%lld\n", query(1, s, 1, n - 1, l, r));
        break;
    }
  }
  return 0;
}
