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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
lli n, tn, node[1 << 20];
bool swapper[20];
void upd(int v, int l, int r, int q, lli val, int dep = n) {
  if (l == r) {
    node[v] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (q <= mid)
    upd(v << 1 | (swapper[dep]), l, mid, q, val, dep - 1);
  else
    upd(v << 1 | (!swapper[dep]), mid + 1, r, q, val, dep - 1);
  node[v] = node[v << 1] + node[v << 1 | 1];
}
lli query(int v, int l, int r, int ql, int qr, int dep = n) {
  if (r < ql || l > qr) return 0;
  if (ql <= l && r <= qr) return node[v];
  int mid = (l + r) / 2;
  return query(v << 1 | (swapper[dep]), l, mid, ql, qr, dep - 1) +
         query(v << 1 | (!swapper[dep]), mid + 1, r, ql, qr, dep - 1);
}
int tejas_919(int kkkk) {
  lli m, k, q, u, v, temp = 0, ans = 0;
  cin >> n >> q;
  tn = (1 << n) - 1;
  for (int i = 0; i < 1 << n; i++) {
    cin >> u;
    upd(1, 0, tn, i, u);
  }
  for (int i = 0; i < q; i++) {
    int tp, x, y;
    cin >> tp >> x;
    if (tp == 1) {
      cin >> y;
      x--;
      upd(1, 0, tn, x, y);
    } else if (tp == 2) {
      for (int i = x; i >= 0; i--) swapper[i] ^= 1;
    } else if (tp == 3) {
      swapper[++x] ^= 1;
    } else {
      cin >> y;
      x--;
      y--;
      cout << query(1, 0, tn, x, y) << '\n';
    }
  }
  return 0;
}
signed main() {
  if (!0) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    tejas_919(i + 1);
  }
  {};
}
