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, position;
long long seg[1 << 20];
long long query(int s, int e, int l, int r, int node, int lev) {
  if (position & (1 << lev)) {
    l = s + e - l;
    r = s + e - r;
    swap(l, r);
  }
  if (e < l || r < s) return 0LL;
  if (s == l && e == r) return seg[node];
  int m = (s + e) / 2;
  if (r <= m) return query(s, m, l, r, node * 2, lev - 1);
  if (l > m) return query(m + 1, e, l, r, node * 2 + 1, lev - 1);
  return query(s, m, l, m, node * 2, lev - 1) +
         query(m + 1, e, m + 1, r, node * 2 + 1, lev - 1);
}
int main() {
  scanf("%d %d", &n, &q);
  int sze = 1 << n;
  for (int i = sze; i < 2 * sze; i++) scanf("%lld", &seg[i]);
  for (int i = sze - 1; i > 0; i--) seg[i] = seg[2 * i] + seg[2 * i + 1];
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int x;
      long long k;
      scanf("%d %lld", &x, &k);
      x--;
      int l = 0, r = sze - 1;
      for (int i = n; i > 0; i--) {
        if (position & (1 << i)) x = l + r - x;
        int m = (l + r) / 2;
        if (x <= m)
          r = m;
        else
          l = m + 1;
      }
      x += sze;
      long long diff = k - seg[x];
      while (x) {
        seg[x] += diff;
        x /= 2;
      }
    } else if (t == 2) {
      int k;
      scanf("%d", &k);
      position ^= (1 << k);
    } else if (t == 3) {
      int k;
      scanf("%d", &k);
      position ^= (3 << k);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%lld\n", query(0, sze - 1, l - 1, r - 1, 1, n));
    }
  }
}
