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 read() {
  char ch = getchar();
  int x = 0, pd = 0;
  while (ch < '0' || ch > '9') pd ^= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return pd ? -x : x;
}
const int maxn = 300005;
int n, m, N;
bool cg[22];
long long sum[maxn << 2];
void pushup(int rt) { rt[sum] = (rt << 1)[sum] + (rt << 1 | 1)[sum]; }
void build(int rt, int l, int r) {
  if (l == r) {
    rt[sum] = read();
    return;
  }
  int mid = (l + r) >> 1;
  build((rt << 1), l, mid), build((rt << 1 | 1), mid + 1, r);
  pushup(rt);
}
void update(int rt, int l, int r, int x, int y, int dep) {
  if (l == r) {
    rt[sum] = y;
    return;
  }
  int mid = (l + r) >> 1;
  if (x <= mid)
    update((rt << 1) ^ cg[dep], l, mid, x, y, dep - 1);
  else
    update((rt << 1 | 1) ^ cg[dep], mid + 1, r, x, y, dep - 1);
  pushup(rt);
}
long long query(int rt, int l, int r, int L, int R, int dep) {
  if (L <= l && r <= R) return rt[sum];
  int mid = (l + r) >> 1;
  if (R <= mid) return query((rt << 1) ^ cg[dep], l, mid, L, R, dep - 1);
  if (L > mid) return query((rt << 1 | 1) ^ cg[dep], mid + 1, r, L, R, dep - 1);
  return query((rt << 1) ^ cg[dep], l, mid, L, R, dep - 1) +
         query((rt << 1 | 1) ^ cg[dep], mid + 1, r, L, R, dep - 1);
}
int main() {
  n = read(), m = read();
  N = 1 << n;
  build(1, 1, N);
  while (m--) {
    int opt = read(), x = read(), y;
    if (opt == 2) {
      for (int i = x; i >= 0; i--) cg[i] ^= 1;
    } else if (opt == 3)
      cg[x + 1] ^= 1;
    else {
      y = read();
      if (opt == 1)
        update(1, 1, N, x, y, n);
      else
        printf("%lld\n", query(1, 1, N, x, y, n));
    }
  }
  return 0;
}
