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;
long long arr[(1 << 20)];
int swp[20];
int n, q;
struct TreeNode {
  long long sum;
  TreeNode *left, *right;
  TreeNode() : sum(0), left(nullptr), right(nullptr) {}
  TreeNode(long long val) : sum(val), left(nullptr), right(nullptr) {}
};
TreeNode* Create(int l, int r) {
  if (l == r) {
    TreeNode* node = new TreeNode(arr[l]);
    return node;
  }
  TreeNode* node = new TreeNode();
  int mid = (l + r) / 2;
  node->left = Create(l, mid);
  node->right = Create(mid + 1, r);
  node->sum = node->left->sum + node->right->sum;
  return node;
}
inline void reverseLevel(int k) {
  for (int i = k; i < 20; i++) swp[i] ^= 1;
}
inline void swapLevel(int k) { swp[k] ^= 1; }
void updateValue(TreeNode* node, int start, int end, int lev, int idx,
                 long long val) {
  if (start == end) {
    node->sum = val;
    return;
  }
  int mid = (start + end) / 2;
  if (start <= idx && idx <= mid) {
    if (swp[lev])
      updateValue(node->right, start, mid, lev + 1, idx, val);
    else
      updateValue(node->left, start, mid, lev + 1, idx, val);
  } else {
    if (swp[lev])
      updateValue(node->left, mid + 1, end, lev + 1, idx, val);
    else
      updateValue(node->right, mid + 1, end, lev + 1, idx, val);
  }
  node->sum = node->left->sum + node->right->sum;
}
long long query(TreeNode* node, int start, int end, int l, int r, int lev) {
  if (start > r or end < l or l > r or start > end) return 0;
  if (start >= l && end <= r) return node->sum;
  int mid = (start + end) / 2;
  long long val;
  if (!swp[lev])
    val = query(node->left, start, mid, l, r, lev + 1) +
          query(node->right, mid + 1, end, l, r, lev + 1);
  else
    val = query(node->right, start, mid, l, r, lev + 1) +
          query(node->left, mid + 1, end, l, r, lev + 1);
  return val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < (1 << n); i++) cin >> arr[i];
  TreeNode* root = Create(0, (1 << n) - 1);
  while (q--) {
    int type;
    int l, r;
    int x, k;
    cin >> type;
    switch (type) {
      case 1:
        cin >> x >> k;
        x--;
        updateValue(root, 0, (1 << n) - 1, 0, x, k);
        break;
      case 2:
        cin >> k;
        reverseLevel(n - k);
        break;
      case 3:
        cin >> k;
        swapLevel(n - k - 1);
        break;
      case 4:
        cin >> l >> r;
        l--;
        r--;
        long long ans = query(root, 0, (1 << n) - 1, l, r, 0);
        cout << ans << "\n";
        break;
    }
  }
}
