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 L[2000000];
int R[2000000];
long long tree[2000000];
vector<int> sw[20];
long long a[2000000];
int ls[2000000];
void build(int x, int l, int r) {
  if (l == r)
    tree[x] = a[l];
  else {
    L[x] = x * 2;
    R[x] = x * 2 + 1;
    build(L[x], l, ((l + r) >> 1));
    build(R[x], ((l + r) >> 1) + 1, r);
    tree[x] = tree[L[x]] + tree[R[x]];
  }
}
int T;
long long query(int x, int l, int r, int s, int e, int lev) {
  if (l > e || r < s) return 0;
  if (l >= s && r <= e) return tree[x];
  if (sw[lev].size() && ls[x] != sw[lev].back()) {
    int num =
        sw[lev].end() - upper_bound(sw[lev].begin(), sw[lev].end(), ls[x]);
    num %= 2;
    if (num) swap(L[x], R[x]);
    ls[x] = sw[lev].back();
  }
  return query(L[x], l, ((l + r) >> 1), s, e, lev + 1) +
         query(R[x], ((l + r) >> 1) + 1, r, s, e, lev + 1);
}
void upd(int x, int l, int r, int id, int val, int lev) {
  if (id < l || id > r) return;
  if (l == r)
    tree[x] = val;
  else {
    if (sw[lev].size() && ls[x] != sw[lev].back()) {
      int num =
          sw[lev].end() - upper_bound(sw[lev].begin(), sw[lev].end(), ls[x]);
      num %= 2;
      if (num) swap(L[x], R[x]);
      ls[x] = sw[lev].back();
    }
    if (id <= ((l + r) >> 1))
      upd(L[x], l, ((l + r) >> 1), id, val, lev + 1);
    else
      upd(R[x], ((l + r) >> 1) + 1, r, id, val, lev + 1);
    tree[x] = tree[L[x]] + tree[R[x]];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  int N = n;
  n = (1 << n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  build(1, 0, n - 1);
  while (q--) {
    T++;
    int t;
    cin >> t;
    if (t == 1) {
      int x, k;
      cin >> x >> k;
      x--;
      upd(1, 0, n - 1, x, k, 0);
    } else if (t == 2) {
      int k;
      cin >> k;
      k = N - k;
      for (int i = k; i <= N; i++) sw[i].push_back(T);
    } else if (t == 3) {
      int k;
      cin >> k;
      k = N - k - 1;
      k = max(k, 0);
      sw[k].push_back(T);
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << query(1, 0, n - 1, l, r, 0) << endl;
    }
  }
}
