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;
const int N = 3e5 + 5;
struct tree {
  int l, r;
  long long sum;
} tr[4 * N];
bool flag[20];
void pushup(int p) { tr[p].sum = tr[p << 1].sum + tr[p << 1 | 1].sum; }
void build(int l, int r, int p) {
  tr[p].l = l, tr[p].r = r;
  if (l == r) {
    scanf("%d", &tr[p].sum);
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, p << 1);
  build(mid + 1, r, p << 1 | 1);
  pushup(p);
}
void modify(int l, int r, int x, int k, int dep, int p) {
  if (l == r) {
    tr[p].sum = k;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    modify(l, mid, x, k, dep - 1, p << 1 | flag[dep]);
  else
    modify(mid + 1, r, x, k, dep - 1, p << 1 | flag[dep] ^ 1);
  pushup(p);
}
long long query(int L, int R, int dep, int p, int l, int r) {
  if (l >= L && r <= R) return tr[p].sum;
  int mid = (l + r) >> 1;
  long long res = 0;
  if (L <= mid) res += query(L, R, dep - 1, p << 1 | flag[dep], l, mid);
  if (R > mid) res += query(L, R, dep - 1, p << 1 | flag[dep] ^ 1, mid + 1, r);
  return res;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  build(1, 1 << n, 1);
  while (q--) {
    int opt, x, k;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d", &x, &k);
      modify(1, 1 << n, x, k, n, 1);
    } else if (opt == 2) {
      scanf("%d", &k);
      for (int i = 0; i <= k; i++) flag[i] ^= 1;
    } else if (opt == 3) {
      scanf("%d", &k);
      flag[k + 1] ^= 1;
    } else {
      scanf("%d%d", &x, &k);
      printf("%lld\n", query(x, k, n, 1, 1, 1 << n));
    }
  }
  return 0;
}
