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 = (int)3e5 + 50;
int n, m, q;
int v[N];
int rv[N], sw[N];
int shad[N];
struct Tree {
  long long dat[N * 4];
  void init_dat(int x, int l, int r) {
    if (l == r) {
      dat[x] = v[l];
      return;
    }
    int mid = (l + r) / 2;
    init_dat(2 * x + 1, l, mid);
    init_dat(2 * x + 2, mid + 1, r);
    dat[x] = dat[2 * x + 1] + dat[2 * x + 2];
  }
  void update(int bt, int pos, int x, int l, int r, int val) {
    int mid = (l + r) / 2;
    if (l == r) {
      dat[x] = val;
      return;
    }
    if (rv[bt]) {
      pos = l + r - pos;
    }
    if (sw[bt - 1]) {
      if (pos <= mid)
        pos += (1 << (bt - 1));
      else
        pos -= (1 << (bt - 1));
    }
    if (pos <= mid)
      update(bt - 1, pos, 2 * x + 1, l, mid, val);
    else
      update(bt - 1, pos, 2 * x + 2, mid + 1, r, val);
    dat[x] = dat[2 * x + 1] + dat[2 * x + 2];
  }
  long long ask(int bt, int a, int b, int x, int l, int r) {
    assert(l <= a && b <= r);
    if (a <= l && r <= b) return dat[x];
    if (rv[bt]) {
      int na = l + r - b;
      int nb = l + r - a;
      a = na, b = nb;
    }
    long long res = 0;
    int mid = (l + r) / 2;
    int blen = (1 << (bt - 1));
    if (min(b, mid) >= a) {
      res += sw[bt - 1] ? ask(bt - 1, a + blen, min(b, mid) + blen, 2 * x + 2,
                              mid + 1, r)
                        : ask(bt - 1, a, min(b, mid), 2 * x + 1, l, mid);
    }
    if (max(a, mid + 1) <= b) {
      res +=
          sw[bt - 1]
              ? ask(bt - 1, max(a, mid + 1) - blen, b - blen, 2 * x + 1, l, mid)
              : ask(bt - 1, max(a, mid + 1), b, 2 * x + 2, mid + 1, r);
    }
    return res;
  }
} tree;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> m >> q;
  n = (1 << m);
  for (int i = 0; i < (n); ++i) cin >> v[i], shad[i] = v[i];
  tree.init_dat(0, 0, n - 1);
  while (q--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int pos, val;
      cin >> pos >> val;
      pos--;
      tree.update(m, pos, 0, 0, n - 1, val);
    } else if (tp == 2) {
      int k;
      cin >> k;
      rv[k] ^= 1;
    } else if (tp == 3) {
      int k;
      cin >> k;
      sw[k] ^= 1;
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << tree.ask(m, l, r, 0, 0, n - 1) << '\n';
    }
  }
}
