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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
struct node {
  long long sum = 0;
  bool rev, swp;
  node *left, *right;
  node(long long sum) : sum(sum) {
    rev = swp = 0;
    left = right = NULL;
  }
};
const int N = (1 << 18) + 5;
vector<int> swp(N, 0);
vector<int> rev(N, 0);
vector<int> v;
int n, q;
void build(node *root, int l = 0, int r = (1 << n) - 1) {
  if (l == r) {
    root->sum = v[l];
    return;
  }
  int mid = (l + r) >> 1;
  root->left = new node(0);
  root->right = new node(0);
  build(root->left, l, mid);
  build(root->right, mid + 1, r);
  root->sum = root->left->sum + root->right->sum;
}
void spread(node *root, int l, int r) {
  if (l != r) {
    int len = (r - l + 1);
    assert((1 << int(log2(len))) == len);
    if (swp[len] ^ root->swp) {
      root->swp ^= 1;
      swap(root->left, root->right);
    }
    if (rev[len] ^ root->rev) {
      root->rev ^= 1;
      swap(root->left, root->right);
      root->left->rev ^= 1;
      root->right->rev ^= 1;
    }
  }
}
void update(int x, int val, node *root, int l = 0, int r = (1 << n) - 1) {
  spread(root, l, r);
  if (l == r) {
    root->sum = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    update(x, val, root->left, l, mid);
  else
    update(x, val, root->right, mid + 1, r);
  root->sum = root->left->sum + root->right->sum;
}
long long query(int x, int y, node *root, int l = 0, int r = (1 << n) - 1) {
  spread(root, l, r);
  if (l > y || r < x || x > y) return 0;
  if (x <= l && r <= y) return root->sum;
  int mid = (l + r) >> 1;
  long long a = query(x, y, root->left, l, mid);
  long long b = query(x, y, root->right, mid + 1, r);
  return a + b;
}
void solve() {
  node *root = new node(0);
  build(root);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int x, k;
      cin >> x >> k;
      update(x - 1, k, root);
    } else if (t == 2) {
      int k;
      cin >> k;
      rev[(1 << k)] ^= 1;
    } else if (t == 3) {
      int k;
      cin >> k;
      if (k == n) continue;
      swp[(1 << (k + 1))] ^= 1;
    } else if (t == 4) {
      int l, r;
      cin >> l >> r;
      cout << query(l - 1, r - 1, root) << "\n";
    }
  }
}
int main() {
  fast();
  cin >> n >> q;
  for (int i = 0; i < (1 << n); i++) {
    int a;
    cin >> a;
    v.push_back(a);
  }
  solve();
}
