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;
long long a[300000];
long long tree[300000 * 4];
bool con[20];
int n;
void init(int idx, int l, int r) {
  if (l == r) {
    tree[idx] = a[l];
    return;
  }
  init(2 * idx, l, (l + r) / 2);
  init(2 * idx + 1, (l + r) / 2 + 1, r);
  tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
}
void upd(int idx, int l, int r, int ini, long long val) {
  if (ini > r || ini < l) {
    return;
  }
  if (l == r) {
    tree[idx] = val;
    return;
  }
  upd(2 * idx, l, (l + r) / 2, ini, val);
  upd(2 * idx + 1, (l + r) / 2 + 1, r, ini, val);
  tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
}
void repl(int pos, int val) {
  int bawa = pos;
  for (int i = 1; i < 20; i++) {
    int now = (1 << i);
    if (con[i]) {
      int x = (bawa - 1) / now + 1;
      int y = (x - 1) * now + 1;
      x = x * now;
      x = x + y;
      bawa = x - bawa;
    }
  }
  upd(1, 1, (1 << n), bawa, val);
}
long long cari(int idx, int l, int r, int kiri, int kanan, int now) {
  if (kiri > r || kanan < l) {
    return 0;
  }
  if (kiri < l) kiri = l;
  if (kanan > r) kanan = r;
  if (con[now]) {
    int x = r + l;
    int p = x - kiri;
    int q = x - kanan;
    kiri = q;
    kanan = p;
  }
  if (l >= kiri && r <= kanan) {
    return tree[idx];
  }
  long long a = cari(2 * idx, l, (l + r) / 2, kiri, kanan, now - 1);
  long long b = cari(2 * idx + 1, (l + r) / 2 + 1, r, kiri, kanan, now - 1);
  return a + b;
}
int main() {
  int q;
  cin >> n >> q;
  int pan = (1 << n);
  memset(con, false, sizeof(con));
  for (int i = 1; i <= pan; i++) {
    cin >> a[i];
  }
  init(1, 1, (1 << n));
  for (int i = 1; i <= q; i++) {
    int tipe;
    cin >> tipe;
    if (tipe == 1) {
      int x, k;
      cin >> x >> k;
      repl(x, k);
    } else if (tipe == 2) {
      int k;
      cin >> k;
      con[k] = !con[k];
    } else if (tipe == 3) {
      int k;
      cin >> k;
      con[k] = !con[k];
      con[k + 1] = !con[k + 1];
    } else {
      int l, r;
      cin >> l >> r;
      long long hasil = cari(1, 1, (1 << n), l, r, n);
      cout << hasil << "\n";
    }
  }
}
