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;
template <typename _T>
inline void read(_T &f) {
  f = 0;
  _T 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 & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
const int N = 1 << 19 | 5;
long long sum[N];
int a[N], tag1[25], tag2[25];
int n, q;
void build(int u, int l, int r) {
  if (l == r) {
    sum[u] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(u << 1, l, mid);
  build(u << 1 | 1, mid + 1, r);
  sum[u] = sum[u << 1] + sum[u << 1 | 1];
}
void change(int u, int l, int r, int x, int y, int dep) {
  if (l == r) {
    sum[u] = y;
    return;
  }
  int mid = (l + r) >> 1;
  if (tag1[dep - 1]) {
    if (mid >= x)
      x += (1 << (dep - 1));
    else
      x -= (1 << (dep - 1));
  }
  if (tag2[dep]) x = l + (1 << dep) - 1 - (x - l);
  if (mid >= x)
    change(u << 1, l, mid, x, y, dep - 1);
  else
    change(u << 1 | 1, mid + 1, r, x, y, dep - 1);
  sum[u] = sum[u << 1] + sum[u << 1 | 1];
}
long long query(int u, int L, int R, int l, int r, int dep) {
  if (l <= L && R <= r) return sum[u];
  int mid = (L + R) >> 1;
  long long ans = 0;
  if (tag2[dep])
    l = L + (1 << dep) - 1 - (l - L), r = L + (1 << dep) - 1 - (r - L),
    swap(l, r);
  if (mid >= l) {
    if (tag1[dep - 1])
      ans += query(u << 1 | 1, mid + 1, R, l + (1 << (dep - 1)),
                   min(r + (1 << (dep - 1)), R), dep - 1);
    else
      ans += query(u << 1, L, mid, l, min(r, mid), dep - 1);
  }
  if (mid + 1 <= r) {
    if (tag1[dep - 1])
      ans += query(u << 1, L, mid, max(L, l - (1 << (dep - 1))),
                   r - (1 << (dep - 1)), dep - 1);
    else
      ans += query(u << 1 | 1, mid + 1, R, max(l, mid + 1), r, dep - 1);
  }
  return ans;
}
int main() {
  read(n);
  read(q);
  for (int i = 1; i <= (1 << n); i++) read(a[i]);
  build(1, 1, 1 << n);
  for (int i = 1; i <= q; i++) {
    int opt;
    read(opt);
    if (opt == 1) {
      int x, k;
      read(x);
      read(k);
      change(1, 1, 1 << n, x, k, n);
    }
    if (opt == 2) {
      int k;
      read(k);
      tag2[k] ^= 1;
    }
    if (opt == 3) {
      int k;
      read(k);
      tag1[k] ^= 1;
    }
    if (opt == 4) {
      int l, r;
      read(l);
      read(r);
      print(query(1, 1, 1 << n, l, r, n), '\n');
    }
  }
  return 0;
}
