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>
const int N = 303000;
using namespace std;
int n, q;
int a[N], s;
long long t[N];
void upd(int x, int y) {
  x += 1;
  while (x < N) {
    t[x] += y;
    x += x & -x;
  }
}
long long get(int x) {
  x += 1;
  long long res = 0;
  while (x > 0) {
    res += t[x];
    x -= x & -x;
  }
  return res;
}
long long solve(int l, int r) {
  if (l == r) {
    return a[l ^ s];
  }
  int h = -1;
  for (int i = 30; i >= 0; i--) {
    int x = (l >> i) & 1;
    int y = (r >> i) & 1;
    if (x != y) {
      h = i;
      break;
    }
  }
  int m = 0;
  for (int i = 30; i > h; i--) {
    int x = (l >> i) & 1;
    m += (x << i);
  }
  if (r - l + 1 == (1 << h + 1)) {
    m ^= ((s >> h + 1) << h + 1);
    return get(m + (1 << h + 1) - 1) - get(m - 1);
  }
  m += (1 << h) - 1;
  return solve(l, m) + solve(m + 1, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 0; i < (1 << n); i++) {
    cin >> a[i];
    upd(i, a[i]);
  }
  for (int i = 1; i <= q; i++) {
    int t, x, y;
    cin >> t;
    if (t == 1) {
      cin >> x >> y;
      x -= 1;
      x ^= s;
      upd(x, y - a[x]);
      a[x] = y;
    } else if (t == 2) {
      cin >> y;
      s ^= (1 << y) - 1;
    } else if (t == 3) {
      cin >> y;
      s ^= (1 << y);
    } else {
      cin >> x >> y;
      x -= 1;
      y -= 1;
      cout << solve(x, y) << "\n";
    }
  }
}
