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 int tree[1000009];
long long int child[1000009];
long long int rev[20], swa[20];
pair<long long int, long long int> children(long long int v, long long int d) {
  pair<long long int, long long int> res = {2 * v, (2 * v) + 1};
  if (swa[d] ^ rev[d]) swap(res.first, res.second);
  return res;
}
void replace(long long int v, long long int tl, long long int tr,
             long long int val, long long int pos, long long int d) {
  if (d == 0) {
    tree[v] = val;
    return;
  }
  long long int mid = (tl + tr) / 2;
  pair<long long int, long long int> ch = children(v, d);
  if (pos <= mid)
    replace(ch.first, tl, mid, val, pos, d - 1);
  else
    replace(ch.second, mid + 1, tr, val, pos, d - 1);
  tree[v] = tree[(2 * v)] + tree[(2 * v) + 1];
}
void reverse(long long int d) {
  for (long long int i = 0; i <= d; i++) {
    rev[i] ^= 1;
  }
}
void swaping(long long int d) { swa[d + 1] ^= 1; }
long long int query(long long int v, long long int tl, long long int tr,
                    long long int l, long long int r, long long int d) {
  if (l > r) return 0;
  if (l == tl && r == tr) return tree[v];
  long long int mid = (tl + tr) / 2;
  pair<long long int, long long int> ch = children(v, d);
  return query(ch.first, tl, mid, l, min(mid, r), d - 1) +
         query(ch.second, mid + 1, tr, max(l, mid + 1), r, d - 1);
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
void solve() {
  long long int n, q;
  cin >> n >> q;
  long long int len = power(2, n);
  long long int x;
  for (long long int i = 0; i < len; i++) {
    cin >> x;
    replace(1, 1, len, x, i + 1, n);
  }
  long long int y, z;
  for (long long int i = 0; i < q; i++) {
    cin >> x;
    if (x == 1) {
      cin >> y >> z;
      replace(1, 1, len, z, y, n);
    } else if (x == 2) {
      cin >> y;
      reverse(y);
    } else if (x == 3) {
      cin >> y;
      swaping(y);
    } else {
      cin >> y >> z;
      cout << query(1, 1, len, y, z, n) << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int begtime = clock();
  long long int t = 1;
  for (long long int i = 0; i < t; i++) {
    solve();
  }
  int endtime = clock();
  cerr << "\n\n"
       << "Time elapsed: " << (endtime - begtime) * 1000 / CLOCKS_PER_SEC
       << " ms\n\n";
  return 0;
}
