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 int N = 1e6 + 6;
const long long MOD = 998244353;
long long a[N], sg[4 * N];
int pox;
void build(int v, int tl, int tr) {
  if (tl == tr) {
    sg[v] = a[tl];
    return;
  }
  int tm = (tl + tr) / 2;
  build(v + v, tl, tm);
  build(v + v + 1, tm + 1, tr);
  sg[v] = sg[v + v] + sg[v + v + 1];
}
void update(int v, int tl, int tr, int ind, long long val) {
  if (tl == tr) {
    sg[v] = val;
    return;
  }
  int tm = (tl + tr) / 2;
  if (ind <= tm)
    update(v + v, tl, tm, ind, val);
  else
    update(v + v + 1, tm + 1, tr, ind, val);
  sg[v] = sg[v + v] + sg[v + v + 1];
}
long long query(int v, int tl, int tr, int ind, int tv) {
  if (tl == tr) {
    return sg[v];
  }
  int tm = (tl + tr) / 2;
  if (pox & (1 << tv)) {
    if (ind & (1 << tv)) {
      return sg[v + v + 1] + query(v + v, tl, tm, ind, tv - 1);
    }
    return query(v + v + 1, tm + 1, tr, ind, tv - 1);
  } else {
    if (ind & (1 << tv)) {
      return sg[v + v] + query(v + v + 1, tm + 1, tr, ind, tv - 1);
    }
    return query(v + v, tl, tm, ind, tv - 1);
  }
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  int qan = n;
  n = (1 << n);
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
  }
  build(1, 0, n - 1);
  for (int i = 0; i < (q); ++i) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int k, x;
      cin >> k >> x;
      k--;
      k ^= pox;
      update(1, 0, n - 1, k, x);
    }
    if (tp == 2) {
      int k;
      cin >> k;
      pox ^= (1 << k) - 1;
    }
    if (tp == 3) {
      int k;
      cin >> k;
      pox ^= (1 << k);
    }
    if (tp == 4) {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      long long ans = query(1, 0, n - 1, r, qan - 1);
      if (l) ans -= query(1, 0, n - 1, l - 1, qan - 1);
      cout << ans << endl;
    }
  }
  return 0;
}
