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 n, m, a[270000], rev[20], l[20], r[20], len[20];
long long c[270000];
void U(int x, int k) {
  x++;
  while (x <= (1 << n)) {
    c[x] += k;
    x += x & -x;
  }
}
long long Q(int x) {
  x++;
  long long r = 0;
  while (x) {
    r += c[x];
    x -= x & -x;
  }
  return r;
}
long long Query(int x) {
  long long ret = 0;
  int cc = 0;
  if (x < 0) return 0;
  for (int i = n, nl = 0; i >= 0; i--)
    if ((x + 1) & (1 << i)) {
      l[++cc] = nl;
      r[cc] = nl + (1 << i) - 1;
      len[cc] = i;
      nl += (1 << i);
    }
  for (int i = 1; i <= n; i++) {
    if (!rev[i]) continue;
    for (int j = 1; j <= cc; j++)
      if (len[j] < i) {
        int L = l[j] % (1 << i), R = r[j] % (1 << i), FI = ((l[j] >> i) << i);
        l[j] = (1 << i) - R - 1 + FI;
        r[j] = (1 << i) - L - 1 + FI;
      }
  }
  for (int i = 1; i <= cc; i++) ret += Q(r[i]) - Q(l[i] - 1);
  return ret;
}
int Find(int x) {
  for (int i = 1; i <= n; i++) {
    if (!rev[i]) continue;
    int FI = ((x >> i) << i);
    x = (1 << i) - x % (1 << i) - 1 + FI;
  }
  return x;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < (1 << n); i++) {
    scanf("%d", &a[i]);
    U(i, a[i]);
  }
  for (int i = 1, opt, x, y; i <= m; i++) {
    scanf("%d%d", &opt, &x);
    if (opt == 1) {
      x--;
      scanf("%d", &y);
      x = Find(x);
      U(x, y - a[x]);
      a[x] = y;
    }
    if (opt == 2) rev[x] ^= 1;
    if (opt == 3) {
      rev[x] ^= 1;
      rev[x + 1] ^= 1;
    }
    if (opt == 4) {
      x--;
      scanf("%d", &y);
      y--;
      printf("%lld\n", Query(y) - Query(x - 1));
    }
  }
  return 0;
}
