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;
int a[(1 << 18)];
long long seg[(1 << 18) * 4];
void init(int pos, int l, int r) {
  if (l == r) {
    seg[pos] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  init(pos * 2, l, mid);
  init(pos * 2 + 1, mid + 1, r);
  seg[pos] = seg[pos * 2] + seg[pos * 2 + 1];
}
void update(int pos, int l, int r, int ind, int val) {
  if (l == r) {
    seg[pos] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (ind <= mid)
    update(pos * 2, l, mid, ind, val);
  else
    update(pos * 2 + 1, mid + 1, r, ind, val);
  seg[pos] = seg[pos * 2] + seg[pos * 2 + 1];
}
long long query(int pos, int l, int r, int ql, int qr) {
  if (r < ql || qr < l) {
    return 0LL;
  }
  if (ql <= l && r <= qr) {
    return seg[pos];
  }
  int mid = (l + r) / 2;
  return query(pos * 2, l, mid, ql, qr) +
         query(pos * 2 + 1, mid + 1, r, ql, qr);
}
long long getSum(int totalBit, int n, int b, int l) {
  long long ret = 0;
  if (l == n) {
    return query(1, 0, n - 1, 0, n - 1);
  }
  int cur = 0, curBit = 0;
  for (int bit = b - 1; bit >= 0; --bit) {
    if (totalBit & (1 << bit)) {
      curBit ^= (1 << bit);
    }
    if (l & (1 << bit)) {
      int ncur = cur ^ curBit;
      int ed = ncur + (1 << bit) - 1;
      assert(ed < n);
      ret += query(1, 0, n - 1, ncur, ed);
      cur += (1 << bit);
    }
  }
  return ret;
}
void solve() {
  int b, q;
  scanf("%d %d ", &b, &q);
  int n = (1 << b);
  for (int i = 0; i < n; ++i) {
    scanf("%d ", &a[i]);
  }
  int bit = 0;
  init(1, 0, n - 1);
  while (q--) {
    int type;
    scanf("%d ", &type);
    int x, y;
    if (type == 1) {
      scanf("%d %d", &x, &y);
      --x;
      x ^= bit;
      update(1, 0, n - 1, x, y);
    } else if (type == 2) {
      scanf("%d ", &x);
      int g = (1 << x) - 1;
      bit ^= g;
    } else if (type == 3) {
      scanf("%d ", &x);
      bit ^= (1 << x);
    } else if (type == 4) {
      scanf("%d %d", &x, &y);
      --x, --y;
      long long ret = getSum(bit, n, b, y + 1);
      if (x - 1 >= 0) {
        ret -= getSum(bit, n, b, x);
      }
      printf("%lld\n", ret);
    } else {
      assert(0);
    }
  }
}
int main() {
  solve();
  return 0;
}
