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>
int Read() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + (c ^ 48), c = getchar();
  return x;
}
const int MAXN = 18;
int N, M, Q, A[(1 << MAXN) + 5];
int Rev[MAXN + 5];
long long Sum[4 * (1 << MAXN) + 5];
void PushUp(int u, int d) {
  Sum[u] = Sum[((u << 1) | Rev[d])] + Sum[((u << 1) | (Rev[d] ^ 1))];
}
void Build(int u, int l, int r, int d) {
  if (l == r) {
    Sum[u] = A[l];
    return;
  }
  int mid = (l + r) >> 1;
  Build(((u << 1) | Rev[d]), l, mid, d + 1);
  Build(((u << 1) | (Rev[d] ^ 1)), mid + 1, r, d + 1);
  PushUp(u, d);
}
void Modify(int u, int l, int r, int p, int val, int d) {
  if (l == r) {
    Sum[u] = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (p <= mid)
    Modify(((u << 1) | Rev[d]), l, mid, p, val, d + 1);
  else
    Modify(((u << 1) | (Rev[d] ^ 1)), mid + 1, r, p, val, d + 1);
  PushUp(u, d);
}
long long Query(int u, int l, int r, int lft, int rgt, int d) {
  if (lft <= l && r <= rgt) return Sum[u];
  int mid = (l + r) >> 1;
  long long ret = 0;
  if (lft <= mid) ret = Query(((u << 1) | Rev[d]), l, mid, lft, rgt, d + 1);
  if (rgt >= mid + 1)
    ret += Query(((u << 1) | (Rev[d] ^ 1)), mid + 1, r, lft, rgt, d + 1);
  return ret;
}
int main() {
  N = 1 << (M = Read()), Q = Read();
  for (int i = 1; i <= N; i++) A[i] = Read();
  Build(1, 1, N, 0);
  while (Q--) {
    int opt = Read();
    if (opt == 1) {
      int x = Read(), k = Read();
      Modify(1, 1, N, x, k, 0);
    } else if (opt == 2) {
      int k = Read();
      for (int i = 1; i <= k; i++) Rev[M - i] ^= 1;
    } else if (opt == 3) {
      int k = Read();
      Rev[M - k - 1] ^= 1;
    } else {
      int l = Read(), r = Read();
      printf("%lld\n", Query(1, 1, N, l, r, 0));
    }
  }
  return 0;
}
