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 INF = 1e9 + 7;
const int mod = 1e9 + 7;
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
class FenwickTree {
 private:
  vector<long long int> ft;

 public:
  FenwickTree() {}
  FenwickTree(int n) { ft.assign(n + 1, 0); }
  long long int rsq(int b) {
    long long int sum = 0;
    for (; b; b -= (b & (-b))) sum += ft[b];
    return sum;
  }
  long long int rsq(int a, int b) { return rsq(b) - (a == 1 ? 0 : rsq(a - 1)); }
  void adjust(int k, int v) {
    for (; k < (int)ft.size(); k += (k & (-k))) ft[k] += v;
  }
};
int s[19];
FenwickTree ft(1 << 19);
vector<int> actual((1 << 19) + 1);
int n, z;
long long int calc(int l, int r, int a, int b) {
  if (r < a || l > b) return 0;
  if (l >= a && r <= b) {
    int diff = r - l;
    for (int j = n - 1; j >= 0 && 1 << j >= diff + 1; j--) {
      if (s[j]) l ^= 1 << j;
    }
    r = l + diff;
    l++, r++;
    return ft.rsq(l, r);
  }
  int mid = (l + r) / 2;
  return calc(l, mid, a, b) + calc(mid + 1, r, a, b);
}
int main() {
  n = nxt(), z = nxt();
  for (int i = 0; i < 1 << n; i++) {
    actual[i + 1] = nxt();
    ft.adjust(i + 1, actual[i + 1]);
  }
  for (int _ = 0; _ < z; _++) {
    int q = nxt();
    if (q == 1) {
      int x = nxt(), k = nxt();
      x--;
      for (int j = n - 1; j >= 0; j--) {
        if (s[j]) {
          x ^= 1 << j;
        }
      }
      x++;
      int diff = k - actual[x];
      ft.adjust(x, diff);
      actual[x] = k;
    } else if (q == 2) {
      int k = nxt();
      for (int i = 0; i < k; i++) s[i] = 1 - s[i];
    } else if (q == 3) {
      int k = nxt();
      s[k] = 1 - s[k];
    } else {
      int l = nxt(), r = nxt();
      l--, r--;
      printf("%lld\n", calc(0, (1 << n) - 1, l, r));
    }
  }
  return 0;
}
