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 tp, q, N;
vector<int> swapped(20);
class SegmentTree {
 private:
  vector<long long> tree;
  vector<long long> raw;
  int n;
  vector<vector<int> > ep;
  long long bound = 0;
  long long merge(long long a, long long b) { return a + b; }
  void buildTree(int l, int r, int p, int k) {
    if (l == r) {
      tree[p] = raw[l];
      return;
    }
    int mid = (l + r) / 2;
    int c1 = 2 * p + 1, c2 = 2 * p + 2;
    ep[c1] = {l, mid};
    ep[c2] = {mid + 1, r};
    buildTree(l, mid, c1, k - 1);
    buildTree(mid + 1, r, c2, k - 1);
    tree[p] = merge(tree[c1], tree[c2]);
  }
  void update(int i, long long val, int l, int r, int p, int k) {
    if (l > i or r < i) return;
    if (l == r) {
      tree[p] = val;
      return;
    }
    int mid = (l + r) / 2;
    int c1 = 2 * p + 1, c2 = 2 * p + 2;
    if (swapped[k]) swap(c1, c2);
    update(i, val, l, mid, c1, k - 1);
    update(i, val, mid + 1, r, c2, k - 1);
    tree[p] = merge(tree[c1], tree[c2]);
  }
  long long sum(int i, int j, int l, int r, int p, int k) {
    if (l > j or r < i) return bound;
    if (l >= i and r <= j) return tree[p];
    int mid = (l + r) / 2;
    int c1 = 2 * p + 1, c2 = 2 * p + 2;
    if (swapped[k]) swap(c1, c2);
    return merge(sum(i, j, l, mid, c1, k - 1),
                 sum(i, j, mid + 1, r, c2, k - 1));
  }

 public:
  SegmentTree(vector<long long> input) {
    raw = input;
    n = raw.size();
    tree.resize(4 * n);
    ep.resize(4 * n, vector<int>(2));
    ep[0] = {0, n - 1};
    buildTree(0, n - 1, 0, tp);
  }
  long long sum(int i, int j) { return sum(i, j, 0, n - 1, 0, tp); }
  void update(int i, long long val) { update(i, val, 0, n - 1, 0, tp); }
  void treeSwap(int k) {
    k += 1;
    swapped[k] = 1 - swapped[k];
  }
  void rev(int k) {
    for (int i = 0; i < (int)k; i++) treeSwap(i);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> tp >> q;
  N = 1 << tp;
  vector<long long> nums(N);
  for (int i = 0; i < (int)N; i++) cin >> nums[i];
  SegmentTree tree(nums);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      tree.update(x - 1, k);
    } else if (t == 2) {
      int k;
      cin >> k;
      tree.rev(k);
    } else if (t == 3) {
      int k;
      cin >> k;
      tree.treeSwap(k);
    } else {
      int l, r;
      cin >> l >> r;
      cout << tree.sum(l - 1, r - 1) << '\n';
    }
  }
  return 0;
}
