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 = (1 << 18) + 1;
struct Node {
  long long val;
  int lef;
  int rig;
  int mask;
};
Node T[N * 2 + 512];
void push(int node, int cl, int cr) {
  int len = (cr - cl + 1);
  if ((T[node].mask & len) == len) swap(T[node].lef, T[node].rig);
  if (cl != cr) {
    T[T[node].lef].mask ^= T[node].mask;
    T[T[node].rig].mask ^= T[node].mask;
  }
  T[node].mask = 0;
}
void upd(int node, int cl, int cr, int pos, int v) {
  push(node, cl, cr);
  if (cl == cr) {
    T[node].val = v;
    return;
  }
  int mid = (cl + cr) / 2;
  if (mid >= pos)
    upd(T[node].lef, cl, mid, pos, v);
  else
    upd(T[node].rig, mid + 1, cr, pos, v);
  T[node].val = T[T[node].lef].val + T[T[node].rig].val;
}
long long get(int node, int cl, int cr, int tl, int tr) {
  push(node, cl, cr);
  if (cr < tl || cl > tr) return 0;
  if (cl >= tl && cr <= tr) {
    return T[node].val;
  }
  int mid = (cl + cr) / 2;
  return get(T[node].lef, cl, mid, tl, tr) +
         get(T[node].rig, mid + 1, cr, tl, tr);
}
int in[N];
void build(int node, int cl, int cr) {
  if (cl == cr) {
    T[node].val = in[cl];
    return;
  }
  T[node].lef = node * 2;
  T[node].rig = node * 2 + 1;
  int mid = (cl + cr) / 2;
  build(node * 2, cl, mid);
  build(node * 2 + 1, mid + 1, cr);
  T[node].val = T[node * 2].val + T[node * 2 + 1].val;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  int m = (1 << n);
  for (int i = 1; i <= m; i++) {
    cin >> in[i];
  }
  build(1, 1, m);
  int typ, l, r;
  int k;
  for (int tc = 1; tc <= q; tc++) {
    cin >> typ;
    if (typ == 1) {
      cin >> l >> r;
      upd(1, 1, m, l, r);
    } else if (typ == 2) {
      cin >> k;
      for (int i = k; i >= 1; i--) {
        T[1].mask ^= (1 << i);
      }
    } else if (typ == 3) {
      cin >> k;
      T[1].mask ^= (1 << (k + 1));
    } else {
      cin >> l >> r;
      cout << get(1, 1, m, l, r) << "\n";
    }
  }
  return 0;
}
