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;
struct seg {
  int k;
  bool rev[19]{};
  long long v = 0;
  seg *ch[2]{};
  seg(int _k) : k(_k) {
    if (k) ch[0] = new seg(k - 1), ch[1] = new seg(k - 1);
  }
  void replace(int l, int r, int i, int d) {
    if (!k) {
      v = d;
      return;
    }
    push();
    int m = l + r >> 1;
    if (i < m)
      ch[0]->replace(l, m, i, d);
    else
      ch[1]->replace(m, r, i, d);
    pull();
  }
  long long query(int l, int r, int _l, int _r) {
    if (_l <= l && r <= _r) return v;
    push();
    long long ans = 0;
    int m = l + r >> 1;
    if (_l < m) ans += ch[0]->query(l, m, _l, _r);
    if (_r > m) ans += ch[1]->query(m, r, _l, _r);
    return ans;
  }
  void pull() { v = ch[0]->v + ch[1]->v; }
  void push() {
    if (rev[k]) swap(ch[0], ch[1]), rev[k] = false;
    for (int i = 0; i < k; ++i)
      ch[0]->rev[i] ^= rev[i], ch[1]->rev[i] ^= rev[i], rev[i] = false;
  }
};
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, q;
  cin >> n >> q;
  int N = 1 << n;
  seg rt(n);
  for (int i = 0; i < (1 << n); ++i) {
    int x;
    cin >> x;
    rt.replace(0, N, i, x);
  }
  while (q--) {
    int ty, x;
    cin >> ty >> x;
    if (ty == 1) {
      int y;
      cin >> y;
      rt.replace(0, N, x - 1, y);
    } else if (ty == 2)
      for (int i = 1; i <= x; ++i) rt.rev[i] ^= true;
    else if (ty == 3)
      rt.rev[x + 1] ^= true;
    else {
      int y;
      cin >> y;
      cout << rt.query(0, N, x - 1, y) << '\n';
    }
  }
}
