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 = (1 << 18) + 100, LG = 20;
int n, swp[LG], lg[N << 2], rv[LG];
long long a[N];
struct Node {
  int lft = 0, rgh = 0, cnt = 0, cnt2 = 0;
  long long sum = 0;
  void init(long long x) { sum = x; }
  void merge(Node L, Node R) { sum = L.sum + R.sum; }
} seg[N << 2];
void shift_stuff(int root) {
  int h = lg[root];
  if (swp[h] % 2 != seg[root].cnt % 2) {
    swap(seg[root].lft, seg[root].rgh);
    seg[root].cnt++;
  }
  if (rv[h] % 2 != seg[root].cnt2 % 2) {
    swap(seg[root].lft, seg[root].rgh);
    seg[root].cnt2++;
    seg[seg[root].lft].cnt2++, seg[seg[root].rgh].cnt2++;
  }
}
void build(int l = 0, int r = n, int root = 1) {
  if (l + 1 == r) {
    seg[root].init(a[l]);
    return;
  }
  int mid = l + r >> 1;
  seg[root].lft = root << 1, seg[root].rgh = root << 1 | 1;
  build(l, mid, seg[root].lft);
  build(mid, r, seg[root].rgh);
  seg[root].merge(seg[seg[root].lft], seg[seg[root].rgh]);
}
void rep(int x, int k, int l = 0, int r = n, int root = 1) {
  if (l + 1 == r) {
    seg[root].init(k);
    return;
  }
  shift_stuff(root);
  int mid = l + r >> 1;
  if (x < mid)
    rep(x, k, l, mid, seg[root].lft);
  else
    rep(x, k, mid, r, seg[root].rgh);
  seg[root].merge(seg[seg[root].lft], seg[seg[root].rgh]);
}
long long sum(int l, int r, int lx = 0, int rx = n, int root = 1) {
  if (l <= lx && r >= rx) return seg[root].sum;
  if (l >= rx || r <= lx) return 0LL;
  int mid = lx + rx >> 1;
  shift_stuff(root);
  return sum(l, r, lx, mid, seg[root].lft) + sum(l, r, mid, rx, seg[root].rgh);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  for (int i = 2; i < N << 2; i++) lg[i] = lg[i / 2] + 1;
  int Q, act;
  cin >> n >> Q;
  act = n;
  n = 1 << n;
  for (int i = 0; i < n; i++) cin >> a[i];
  build();
  while (Q--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int x, k;
      cin >> x >> k;
      x--;
      rep(x, k);
    } else if (tp == 2) {
      int k;
      cin >> k;
      rv[act - k]++;
    } else if (tp == 3) {
      int k;
      cin >> k;
      swp[act - k - 1]++;
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      cout << sum(l, r) << '\n';
    }
  }
  return 0;
}
