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 n, q, m[1000000], tag[20], seg[1000000 * 4];
long long int a, b, c;
void pull(long long int nd, long long int l) {
  seg[nd] = seg[tag[l + 1] == 1 ? nd * 2 + 1 : nd * 2] +
            seg[tag[l + 1] == 1 ? nd * 2 : nd * 2 + 1];
  return;
}
void build(long long int nd, long long int L, long long int R,
           long long int l) {
  if (L == R)
    seg[nd] = m[L];
  else {
    long long int mid = (L + R) / 2;
    build(tag[l + 1] == 1 ? nd * 2 + 1 : nd * 2, L, mid, l + 1);
    build(tag[l + 1] == 1 ? nd * 2 : nd * 2 + 1, mid + 1, R, l + 1);
    pull(nd, l);
  }
  return;
}
void update(long long int nd, long long int L, long long int R,
            long long int id, long long int val, long long int l) {
  if (L == R) {
    seg[nd] = val;
  } else {
    long long int mid = (L + R) / 2;
    if (id <= mid)
      update(tag[l + 1] == 1 ? nd * 2 + 1 : nd * 2, L, mid, id, val, l + 1);
    else
      update(tag[l + 1] == 1 ? nd * 2 : nd * 2 + 1, mid + 1, R, id, val, l + 1);
    pull(nd, l);
  }
  return;
}
long long int query(long long int nd, long long int L, long long int R,
                    long long int l, long long int r, long long int lay) {
  if (l > R || r < L) return 0;
  if (l <= L && r >= R)
    return seg[nd];
  else {
    long long int mid = (L + R) / 2;
    return query(tag[lay + 1] == 1 ? nd * 2 + 1 : nd * 2, L, mid, l, r,
                 lay + 1) +
           query(tag[lay + 1] == 1 ? nd * 2 : nd * 2 + 1, mid + 1, R, l, r,
                 lay + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= (1 << n); i++) {
    cin >> m[i];
  }
  build(1, 1, (1 << n), 0);
  while (q--) {
    cin >> a;
    if (a == 1) {
      cin >> b >> c;
      update(1, 1, (1 << n), b, c, 0);
    } else if (a == 2) {
      cin >> b;
      for (int i = n - b + 1; i <= n; i++) {
        tag[i] ^= 1;
      }
    } else if (a == 3) {
      cin >> b;
      tag[n - b] ^= 1;
    } else {
      cin >> b >> c;
      cout << query(1, 1, (1 << n), b, c, 0) << "\n";
    }
  }
  return 0;
}
