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;
struct node {
  int lson, rson, depth;
  long long val;
  bool reverse;
} tree[1600010];
int n, q, a[400010], tot;
int pub[20];
int build(int cl, int cr) {
  int x = ++tot;
  if (cl == cr) {
    tree[x].val = a[cl];
  } else {
    int mid = (cl + cr) / 2;
    tree[x].lson = build(cl, mid);
    tree[x].rson = build(mid + 1, cr);
    tree[x].val = tree[tree[x].lson].val + tree[tree[x].rson].val;
    tree[x].depth = tree[tree[x].lson].depth + 1;
  }
  return x;
}
long long query(int x, int l, int r, int L, int R) {
  if (r < L || R < l) return 0;
  if (L <= l && r <= R) return tree[x].val;
  if (tree[x].reverse != pub[tree[x].depth])
    swap(tree[x].lson, tree[x].rson), tree[x].reverse ^= 1;
  int mid = (l + r) / 2;
  return query(tree[x].lson, l, mid, L, R) +
         query(tree[x].rson, mid + 1, r, L, R);
}
void chenge(int x, int l, int r, int pos, long long val) {
  tree[x].val += val;
  if (l == r) return;
  if (tree[x].reverse != pub[tree[x].depth])
    swap(tree[x].lson, tree[x].rson), tree[x].reverse ^= 1;
  int mid = (l + r) / 2;
  if (pos <= mid)
    chenge(tree[x].lson, l, mid, pos, val);
  else
    chenge(tree[x].rson, mid + 1, r, pos, val);
}
int main() {
  scanf("%d%d", &n, &q);
  int tmp = 1 << n;
  for (int i = 1; i <= tmp; i++) scanf("%d", &a[i]);
  build(1, tmp);
  for (int op, x, y, i = 1; i <= q; i++) {
    scanf("%d%d", &op, &x);
    if (op == 1) {
      scanf("%d", &y);
      long long fuck = query(1, 1, tmp, x, x);
      chenge(1, 1, tmp, x, y - fuck);
    }
    if (op == 2) {
      for (int i = x; i > 0; i--) pub[i] ^= 1;
    }
    if (op == 3) {
      pub[x + 1] ^= 1;
    }
    if (op == 4) {
      scanf("%d", &y);
      printf("%lld\n", query(1, 1, tmp, x, y));
    }
  }
  return 0;
}
