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;
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  os << "]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const long long N = 300005;
long long a[N], rem[20], mn, n, q;
struct tr {
  long long val;
  tr *left, *right;
} * root;
void create(tr *root, long long start, long long end) {
  if (start == end) {
    root->val = a[start];
    return;
  }
  long long mid = start + end >> 1;
  root->left = new tr();
  root->right = new tr();
  create(root->left, start, mid);
  create(root->right, mid + 1, end);
  root->val = root->left->val + root->right->val;
}
void update(tr *root, long long start, long long end, long long in,
            long long val, long long level) {
  if (start == end) {
    root->val = val;
    return;
  }
  long long mid = start + end >> 1;
  if (rem[level - 1]) {
    if (in <= mid)
      update(root->right, start, mid, in, val, level - 1);
    else
      update(root->left, mid + 1, end, in, val, level - 1);
  } else {
    if (in <= mid)
      update(root->left, start, mid, in, val, level - 1);
    else
      update(root->right, mid + 1, end, in, val, level - 1);
  }
  root->val = root->left->val + root->right->val;
}
long long query(tr *root, long long start, long long end, long long l,
                long long r, long long level) {
  if (start > r || end < l || start > end) return 0;
  if (start >= l && end <= r) return root->val;
  long long mid = start + end >> 1;
  if (!rem[level - 1]) {
    return query(root->left, start, mid, l, r, level - 1) +
           query(root->right, mid + 1, end, l, r, level - 1);
  }
  return query(root->left, mid + 1, end, l, r, level - 1) +
         query(root->right, start, mid, l, r, level - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  root = new tr();
  for (long long i = 1; i < (1LL << n) + 1; i++) cin >> a[i];
  create(root, 1, (1LL << n));
  while (q--) {
    long long c;
    cin >> c;
    if (c == 1) {
      long long x, k;
      cin >> x >> k;
      update(root, 1, (1LL << n), x, k, n);
    } else if (c == 2) {
      long long k;
      cin >> k;
      for (long long j = k - 1; j >= 0; j--) rem[j] ^= 1;
    } else if (c == 3) {
      long long k;
      cin >> k;
      rem[k] ^= 1;
    } else {
      long long l, r;
      cin >> l >> r;
      long long ans = query(root, 1, (1LL << n), l, r, n);
      cout << ans << '\n';
    }
  }
  return 0;
}
