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;
int n, q;
int mask;
int All[19];
long long arr[1 << 18];
long long SegTree[(1 << 19) | 5];
void Update(int pos, long long val) {
  for (pos = pos + (1 << n); pos > 0; pos >>= 1) {
    SegTree[pos] += val;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  All[0] = 0;
  for (int i = 1; i <= n; i++) {
    All[i] = (All[i - 1] << 1) | 1;
  }
  for (int i = 0; i < 1 << n; i++) {
    cin >> arr[i];
    Update(i, arr[i]);
  }
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int pos, val;
      cin >> pos >> val;
      --pos;
      pos ^= mask;
      Update(pos, val - arr[pos]);
      arr[pos] = val;
    } else if (type == 2) {
      int k;
      cin >> k;
      mask ^= ((1 << k) - 1);
    } else if (type == 3) {
      int k;
      cin >> k;
      mask ^= (1 << k);
    } else {
      int l, r;
      long long ans = 0;
      cin >> l >> r;
      --l;
      --r;
      while (l <= r) {
        for (int sz = n; sz >= 0; sz--) {
          if (!(l & All[sz])) {
            if (l + (1 << sz) - 1 <= r) {
              int realmask = mask >> sz, reall = l >> sz;
              reall ^= realmask;
              ans += SegTree[reall + (1 << (n - sz))];
              l += (1 << sz);
              break;
            }
          }
        }
      }
      cout << ans << '\n';
    }
  }
}
