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;
struct Node {
  long long sm;
  long long cnt_sw;
  long long depth;
  Node *l, *r;
  Node(long long val_)
      : sm(val_), depth(0), cnt_sw(0), l(nullptr), r(nullptr) {}
};
const long long MAXN = 19;
long long a[(1 << MAXN)], swap_cnt[MAXN];
void build(Node *&root, long long l, long long r, long long dep = 0) {
  if (l + 1 == r) {
    root = new Node(a[l]);
    root->depth = dep;
    return;
  }
  root = new Node(0);
  root->depth = dep;
  long long m = (l + r) / 2;
  build(root->l, l, m, dep + 1);
  build(root->r, m, r, dep + 1);
  root->sm = root->l->sm + root->r->sm;
}
void push(Node *root) {
  if (swap_cnt[root->depth] % 2 != root->cnt_sw % 2) {
    swap(root->l, root->r);
    root->cnt_sw++;
  }
}
void change(Node *root, long long l, long long r, long long pos, long long x) {
  push(root);
  if (l + 1 == r) {
    root->sm = x;
    return;
  }
  long long m = (l + r) / 2;
  if (pos < m) {
    change(root->l, l, m, pos, x);
  } else {
    change(root->r, m, r, pos, x);
  }
  root->sm = root->l->sm + root->r->sm;
}
long long get_sm(Node *root, long long l, long long r, long long ql,
                 long long qr) {
  push(root);
  if (l == ql && r == qr) {
    return root->sm;
  }
  long long m = (l + r) / 2;
  if (qr <= m) {
    return get_sm(root->l, l, m, ql, qr);
  }
  if (ql >= m) {
    return get_sm(root->r, m, r, ql, qr);
  }
  return get_sm(root->l, l, m, ql, m) + get_sm(root->r, m, r, m, qr);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < (1 << n); i++) {
    cin >> a[i];
  }
  Node *root = nullptr;
  build(root, 0, 1 << n);
  long long x, k, l, r;
  while (q--) {
    long long op;
    cin >> op;
    if (op == 1) {
      cin >> x >> k;
      change(root, 0, 1 << n, x - 1, k);
    } else if (op == 2) {
      cin >> k;
      k = n - k;
      for (long long i = k; i <= n; i++) {
        swap_cnt[i]++;
      }
    } else if (op == 3) {
      cin >> k;
      swap_cnt[n - k - 1]++;
    } else {
      cin >> l >> r;
      cout << get_sm(root, 0, 1 << n, l - 1, r) << '\n';
    }
  }
}
