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;
const long long N = 1 << 18;
struct segT {
  long long data[4 * N], swp[20];
  void modify(long long pos, long long val, long long l, long long r,
              long long ind = 1, long long dep = 0) {
    if (l == r) {
      data[ind] = val;
    } else {
      long long mid = l + r >> 1;
      if (swp[dep]) {
        if (pos <= mid)
          modify(pos, val, l, mid, ind << 1 | 1, dep + 1);
        else
          modify(pos, val, mid + 1, r, ind << 1, dep + 1);
        data[ind] = data[ind << 1] + data[ind << 1 | 1];
      } else {
        if (pos <= mid)
          modify(pos, val, l, mid, ind << 1, dep + 1);
        else
          modify(pos, val, mid + 1, r, ind << 1 | 1, dep + 1);
        data[ind] = data[ind << 1] + data[ind << 1 | 1];
      }
    }
  }
  long long query(long long ql, long long qr, long long l, long long r,
                  long long ind = 1, long long dep = 0) {
    if (ql > r || qr < l) return 0;
    if (ql <= l && qr >= r) return data[ind];
    long long mid = l + r >> 1;
    if (swp[dep]) {
      return query(ql, qr, l, mid, ind << 1 | 1, dep + 1) +
             query(ql, qr, mid + 1, r, ind << 1, dep + 1);
    } else {
      return query(ql, qr, l, mid, ind << 1, dep + 1) +
             query(ql, qr, mid + 1, r, ind << 1 | 1, dep + 1);
    }
  }
} st;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1, tmp; i <= 1 << n; i++) {
    cin >> tmp;
    st.modify(i, tmp, 1, 1 << n);
  }
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      st.modify(x, k, 1, 1 << n);
    } else if (t == 2) {
      long long k;
      cin >> k;
      for (long long i = n - k; i < n; i++) {
        st.swp[i] ^= 1;
      }
    } else if (t == 3) {
      long long k;
      cin >> k;
      st.swp[n - k - 1] ^= 1;
    } else {
      long long l, r;
      cin >> l >> r;
      cout << st.query(l, r, 1, 1 << n) << "\n";
    }
  }
}
