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;
vector<long long> a;
int n, q;
vector<bool> swap_indx;
vector<bool> reverse_array;
struct S {
  vector<long long> arr;
  S *left, *right;
  long long sum;
  int k;
  S() { sum = 0LL; }
  long long total(int left_q, int right_q) {
    if (right_q < 0 or left_q >= (int)arr.size()) return 0LL;
    if (left_q <= 0 and right_q + 1 >= (int)arr.size()) return sum;
    left_q = max(0, left_q);
    right_q = min(right_q, (int)arr.size() - 1);
    long long suma = 0LL;
    S* L = this->left;
    S* R = this->right;
    if (swap_indx[k]) swap(L, R);
    if (reverse_array[k]) {
      left_q = (int)arr.size() - left_q - 1;
      right_q = (int)arr.size() - right_q - 1;
      swap(left_q, right_q);
    }
    return L->total(left_q, right_q) +
           R->total(left_q - (1 << (k - 1)), right_q - (1 << (k - 1)));
  }
  void update(int pos, long long val) {
    if (pos < 0 or pos >= (int)arr.size()) return;
    S* L = this->left;
    S* R = this->right;
    if (swap_indx[k]) swap(L, R);
    if (reverse_array[k]) {
      pos = (int)arr.size() - pos - 1;
    }
    if (k > 0) {
      L->update(pos, val);
      R->update(pos - (1 << (k - 1)), val);
      sum = L->sum + R->sum;
    } else {
      sum += (val - arr[pos]);
      arr[pos] = val;
    }
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  swap_indx.resize(n + 1, false);
  reverse_array.resize(n + 1, false);
  a.resize((1 << n));
  for (int i = 0; i < a.size(); i++) cin >> a[i];
  vector<vector<S> > weas(n + 1);
  for (int i = 0; i <= n; i++) {
    int tamano = (1 << i);
    int indx = 0;
    while (indx < (1 << n)) {
      weas[i].push_back(S());
      for (int j = 0; j < tamano; j++) {
        weas[i][weas[i].size() - 1].arr.push_back(a[indx + j]);
        weas[i][weas[i].size() - 1].sum += weas[i][weas[i].size() - 1].arr[j];
      }
      if (i > 0) {
        int s = weas[i].size() - 1;
        weas[i][weas[i].size() - 1].left = &weas[i - 1][s << 1];
        weas[i][weas[i].size() - 1].right = &weas[i - 1][s << 1 | 1];
      }
      weas[i][weas[i].size() - 1].k = i;
      indx += tamano;
    }
  }
  while (q--) {
    int t;
    cin >> t;
    if (t == 4) {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << weas[n][0].total(l, r) << endl;
    } else if (t == 3) {
      int K;
      cin >> K;
      swap_indx[K + 1] = !swap_indx[K + 1];
    } else if (t == 2) {
      int K;
      cin >> K;
      reverse_array[K] = !reverse_array[K];
    } else {
      long long x;
      int K;
      cin >> K >> x;
      K--;
      weas[n][0].update(K, x);
    }
  }
  return 0;
}
