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 Xor = 0;
struct fenwick {
  int size;
  vector<long long> p;
  void build(int size_) {
    size = size_;
    p.resize(size);
  }
  fenwick() {}
  fenwick(int size) { build(size); }
  long long get(int l) {
    long long r = 0;
    for (int i = l; i >= 0; i = (i & (i + 1)) - 1) r += p[i];
    return r;
  }
  long long get(int l, int r) { return get(r) - get(l - 1); }
  void upd(int ind, long long val) {
    for (int i = ind; i < size; i |= i + 1) p[i] += val;
  }
  void set_elem(int ind, int val) { upd(ind, val - get(ind, ind)); }
};
fenwick fen;
long long get(int l, int r, int x, int y) {
  if (l >= x && r <= y) {
    int L = (l ^ (Xor & (~(r - l))));
    int R = (r ^ (Xor & (~(r - l))));
    long long vl = fen.get(L, R);
    return vl;
  }
  if (l > y || r < x) return 0;
  int md = r + l >> 1;
  return get(l, md, x, y) + get(md + 1, r, x, y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  fen.build(1 << n);
  for (int i = 0; i < (1 << n); ++i) {
    int a;
    cin >> a;
    fen.upd(i, a);
  }
  for (int i = 0; i < q; ++i) {
    int t, x, k, l, r;
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      x = ((x - 1) ^ Xor);
      fen.set_elem(x, k);
    }
    if (t == 2) {
      cin >> x;
      Xor ^= (1 << x) - 1;
    }
    if (t == 3) {
      cin >> x;
      Xor ^= (1 << x);
    }
    if (t == 4) {
      cin >> l >> r;
      l--, r--;
      cout << get(0, (1 << n) - 1, l, r) << '\n';
    }
  }
}
