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 = 20;
int n, q;
int a[1 << N];
bool flip[N], swapVal[N];
struct IT {
  long long sum[4 << N];
  IT() { memset(sum, 0, sizeof sum); }
  void init(int node, int l, int r) {
    if (l == r) {
      sum[node] = a[l];
      return;
    }
    int m = l + r >> 1;
    init(node << 1, l, m);
    init(node << 1 | 1, m + 1, r);
    sum[node] = sum[node << 1] + sum[node << 1 | 1];
  }
  void update(int node, int x, int k, int lv) {
    if (lv == 0) {
      sum[node] = k;
      return;
    }
    if (flip[lv]) {
      x = (1 << lv) - x - 1;
    }
    if (swapVal[lv]) {
      x ^= (1 << lv - 1);
    }
    if (x >> lv - 1 & 1) {
      update(node << 1 | 1, x ^ (1 << lv - 1), k, lv - 1);
    } else {
      update(node << 1, x, k, lv - 1);
    }
    sum[node] = sum[node << 1] + sum[node << 1 | 1];
  }
  long long get(int node, int x, int y, int lv) {
    if (x == 0 && y == (1 << lv) - 1) {
      return sum[node];
    }
    if (flip[lv]) {
      x = (1 << lv) - x - 1;
      y = (1 << lv) - y - 1;
      swap(x, y);
    }
    int m = (1 << lv - 1);
    long long res = 0;
    if (y < m) {
      res += get(node << 1 | swapVal[lv], x, y, lv - 1);
    } else if (x >= m) {
      res += get(node << 1 | (swapVal[lv] ^ 1), x ^ m, y ^ m, lv - 1);
    } else {
      if (!swapVal[lv]) {
        res += get(node << 1, x, m - 1, lv - 1);
        res += get(node << 1 | 1, 0, y ^ m, lv - 1);
      } else {
        res += get(node << 1, 0, y ^ m, lv - 1);
        res += get(node << 1 | 1, x, m - 1, lv - 1);
      }
    }
    return res;
  }
} it;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < (1 << n); i++) {
    cin >> a[i];
  }
  it.init(1, 0, (1 << n) - 1);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x, k;
      cin >> x >> k;
      x--;
      it.update(1, x, k, n);
    } else if (op == 2) {
      int k;
      cin >> k;
      flip[k] ^= 1;
    } else if (op == 3) {
      int k;
      cin >> k;
      swapVal[k + 1] ^= 1;
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << it.get(1, l, r, n) << '\n';
    }
  }
  return 0;
}
