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;
using ll = long long;
const int maxN = 1 << 19;
int n, q, mask = 0, k, l, r, x;
ll seg[maxN];
void update(int node) {
  if (node == 0) return;
  seg[node] = seg[node << 1] + seg[(node << 1) + 1];
  update(node / 2);
}
ll query(int node, int level, int left, int right) {
  ll ans = 0;
  if (right - left + 1 == (1 << level)) return seg[node];
  int mid = (1 << level);
  mid = (left / mid) * mid + (mid >> 1);
  int sw = (1 << (level - 1));
  if (right < mid) {
    if (mask & (1 << level)) {
      ans += query(2 * node + 1, level - 1, left + sw, right + sw);
    } else
      ans += query(2 * node, level - 1, left, right);
  } else if (left >= mid) {
    if (mask & (1 << level)) {
      ans += query(2 * node, level - 1, left - sw, right - sw);
    } else
      ans += query(2 * node + 1, level - 1, left, right);
  } else {
    if (mask & (1 << level)) {
      ans += query(2 * node, level - 1, mid - sw, right - sw);
      ans += query(2 * node + 1, level - 1, left + sw, mid + sw - 1);
    } else {
      ans += query(2 * node, level - 1, left, mid - 1);
      ans += query(2 * node + 1, level - 1, mid, right);
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = (1 << n); i < (1 << (n + 1)); ++i) cin >> seg[i];
  for (int i = (1 << n) - 1; i > 0; --i) {
    seg[i] = seg[i << 1] + seg[(i << 1) + 1];
  }
  for (int i = 0; i < q; ++i) {
    int type;
    cin >> type;
    if (type == 1) {
      cin >> x >> k;
      --x;
      x ^= (mask >> 1);
      seg[(1 << n) + x] = k;
      update(((1 << n) + x) / 2);
    }
    if (type == 2) {
      cin >> k;
      mask ^= ((1 << (k + 1)) - 1);
    }
    if (type == 3) {
      cin >> k;
      mask ^= (1 << (k + 1));
    }
    if (type == 4) {
      cin >> l >> r;
      --l, --r;
      cout << query(1, n, l, r) << endl;
    }
  }
  return 0;
}
