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;
const long long INF = 2e9;
const long long MB = 20;
const long long MOD = 1e9 + 7;
const long long maxN = 19;
struct node {
  node* left;
  node* right;
  long long cnt, sum, h;
  node(long long x) {
    left = nullptr;
    right = nullptr;
    cnt = 0;
    sum = 0;
    h = x;
  }
};
long long rev[maxN];
void build(node* root, long long l, long long r, long long h,
           vector<long long>& a) {
  if (l + 1 == r) {
    root->sum = a[l];
    return;
  }
  long long m = (l + r) >> 1;
  root->left = new node(h - 1);
  root->right = new node(h - 1);
  build(root->left, l, m, h - 1, a);
  build(root->right, m, r, h - 1, a);
  root->sum = root->left->sum + root->right->sum;
}
void update(node* root, long long l, long long r, long long id, long long val) {
  if (l + 1 == r) {
    root->sum = val;
    return;
  }
  if ((rev[root->h] - root->cnt) % 2) swap(root->left, root->right);
  root->cnt = rev[root->h];
  long long m = (l + r) >> 1;
  if (id < m)
    update(root->left, l, m, id, val);
  else
    update(root->right, m, r, id, val);
  root->sum = root->left->sum + root->right->sum;
}
long long getSum(node* root, long long l, long long r, long long tl,
                 long long tr) {
  if (tl <= l && r <= tr) {
    return root->sum;
  }
  if ((rev[root->h] - root->cnt) % 2) swap(root->left, root->right);
  root->cnt = rev[root->h];
  long long m = (l + r) >> 1;
  long long res = 0;
  if (tl < m) res += getSum(root->left, l, m, tl, tr);
  if (m < tr) res += getSum(root->right, m, r, tl, tr);
  return res;
}
void solve() {
  long long n, q;
  cin >> n >> q;
  vector<long long> a(1 << n);
  for (long long i = 0; i < a.size(); i++) {
    cin >> a[i];
  }
  long long m = 1 << n;
  node* main_root = new node(n);
  build(main_root, 0, m, n, a);
  for (long long i = 0; i < q; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      x--;
      update(main_root, 0, m, x, k);
    } else if (t == 2) {
      long long k;
      cin >> k;
      for (long long i = k; i > 0; i--) rev[i]++;
    } else if (t == 3) {
      long long k;
      cin >> k;
      rev[k + 1]++;
    } else {
      long long l, r;
      cin >> l >> r;
      l--;
      cout << getSum(main_root, 0, m, l, r) << '\n';
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed;
  cout.precision(12);
  long long t = 1;
  while (t--) solve();
}
