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;
inline int qr() {
  int f = 0, fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + c - 48;
    c = getchar();
  }
  return f * fu;
}
const int N = (1 << 18) + 10;
int tag;
struct Seg_Tree {
  struct {
    int l, r;
    long long dat;
  } t[N * 4];
  void build(int p, int l, int r) {
    t[p].l = l, t[p].r = r;
    if (l == r) {
      t[p].dat = qr();
      return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    t[p].dat = t[p << 1].dat + t[p << 1 | 1].dat;
  }
  void change(int p, int x, int v) {
    if (t[p].l == t[p].r) {
      t[p ^ tag].dat = v;
      return;
    }
    int mid = (t[p].l + t[p].r) >> 1, d = log2(t[p].r - t[p].l + 1);
    change(x <= mid ? p << 1 : p << 1 | 1, x, v);
    t[p ^ (tag >> d)].dat = t[(p << 1) ^ (tag >> (d - 1))].dat +
                            t[(p << 1 | 1) ^ (tag >> (d - 1))].dat;
  }
  long long ask(int p, int l, int r) {
    if (l <= t[p].l && r >= t[p].r)
      return t[p ^ (tag >> (int)log2(t[p].r - t[p].l + 1))].dat;
    int mid = (t[p].l + t[p].r) >> 1;
    long long val = 0;
    if (l <= mid) val = ask(p << 1, l, r);
    if (r > mid) val += ask(p << 1 | 1, l, r);
    return val;
  }
} tr;
int n, q, op;
int main() {
  n = qr(), q = qr();
  tr.build(1, 1, 1 << n);
  while (q--) {
    op = qr();
    if (op == 1) {
      int x = qr(), k = qr();
      tr.change(1, x, k);
    } else if (op == 2)
      tag ^= (1 << qr()) - 1;
    else if (op == 3)
      tag ^= 1 << qr();
    else {
      int l = qr(), r = qr();
      printf("%lld\n", tr.ask(1, l, r));
    }
  }
  return 0;
}
