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;
vector<long long> st;
long long n, q;
vector<int> swp(20, false);
vector<long long> arr;
void build(long long ind, long long l, long long r) {
  if (l == r) {
    st[ind] = arr[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(2 * ind, l, mid);
  build(2 * ind + 1, mid + 1, r);
  st[ind] = st[2 * ind] + st[2 * ind + 1];
}
void update(long long ind, long long l, long long r, long long req,
            long long level) {
  if (req < l || req > r) return;
  if (l == r) {
    st[ind] = arr[req];
    return;
  }
  long long mid = (l + r) >> 1;
  if (swp[level - 1]) {
    update(2 * ind + 1, l, mid, req, level - 1);
    update(2 * ind, mid + 1, r, req, level - 1);
  } else {
    update(2 * ind, l, mid, req, level - 1);
    update(2 * ind + 1, mid + 1, r, req, level - 1);
  }
  st[ind] = st[2 * ind + 1] + st[2 * ind];
}
long long query(long long ind, long long l, long long r, long long ql,
                long long qr, long long level) {
  if (l > qr || r < ql) return 0;
  if (ql <= l && r <= qr) return st[ind];
  long long mid = (l + r) / 2;
  if (swp[level - 1])
    return query(2 * ind + 1, l, mid, ql, qr, level - 1) +
           query(2 * ind, mid + 1, r, ql, qr, level - 1);
  else
    return query(2 * ind, l, mid, ql, qr, level - 1) +
           query(2 * ind + 1, mid + 1, r, ql, qr, level - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  long long len = 1 << n;
  arr.resize(len);
  st.assign(4 * len, 0);
  for (long long i = 0; i < len; i++) cin >> arr[i];
  build(1, 0, len - 1);
  while (q--) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long ind, val;
      cin >> ind >> val;
      ind--;
      arr[ind] = val;
      update(1, 0, len - 1, ind, n);
    } else if (type == 2) {
      long long k;
      cin >> k;
      for (long long i = 0; i < k; i++) swp[i] ^= true;
    } else if (type == 3) {
      long long k;
      cin >> k;
      swp[k] ^= true;
    } else {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << query(1, 0, len - 1, l, r, n) << '\n';
    }
  }
  return 0;
}
