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 N = ((1 << 18) + 100);
const int M = N << 2;
long long sum[M], L[M], R[M];
int n, q, Rev[30], Swap[30];
inline void up(int x) {
  sum[x] = sum[(x << 1)] + sum[(x << 1 | 1)];
  return;
}
inline void build(int x, int l, int r) {
  L[x] = l;
  R[x] = r;
  if (l == r) {
    scanf("%lld", &sum[x]);
    return;
  }
  int mid = (l + r) >> 1;
  build((x << 1), l, mid);
  build((x << 1 | 1), mid + 1, r);
  up(x);
}
inline void update(int x, int id, int val, int dep = n) {
  if (L[x] == R[x] && L[x] == id) {
    sum[x] = val;
    return;
  }
  int mid = (L[x] + R[x]) >> 1;
  if (Swap[dep]) {
    if (id <= mid)
      id = id + (1 << (dep - 1));
    else
      id = id - (1 << (dep - 1));
  }
  if (Rev[dep]) id = L[x] + R[x] - id;
  if (id <= mid)
    update((x << 1), id, val, dep - 1);
  else
    update((x << 1 | 1), id, val, dep - 1);
  up(x);
}
long long Query(int x, int l, int r, int dep = n) {
  if (L[x] == l && R[x] == r) return sum[x];
  int mid = (L[x] + R[x]) >> 1, F = 0;
  if (Rev[dep]) {
    l = L[x] + R[x] - l;
    r = L[x] + R[x] - r;
    swap(l, r);
  }
  if (Swap[dep]) {
    if (l <= mid)
      l = l + (1 << (dep - 1));
    else
      l = l - (1 << (dep - 1));
    if (r <= mid)
      r = r + (1 << (dep - 1));
    else
      r = r - (1 << (dep - 1));
    if (l > r) F = 1;
  }
  if (!F) {
    if (r <= mid)
      return Query((x << 1), l, r, dep - 1);
    else if (l >= mid + 1)
      return Query((x << 1 | 1), l, r, dep - 1);
    else
      return Query((x << 1), l, mid, dep - 1) +
             Query((x << 1 | 1), mid + 1, r, dep - 1);
  } else
    return Query((x << 1), L[x], r, dep - 1) +
           Query((x << 1 | 1), l, R[x], dep - 1);
}
int main() {
  scanf("%d%d", &n, &q);
  build(1, 1, (1 << n));
  while (q--) {
    int op, x, k, l, r;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &x, &k);
      update(1, x, k);
    }
    if (op == 2) {
      scanf("%d", &k);
      Rev[k] ^= 1;
    }
    if (op == 3) {
      scanf("%d", &k);
      Swap[k + 1] ^= 1;
    }
    if (op == 4) {
      scanf("%d%d", &l, &r);
      printf("%lld\n", Query(1, l, r));
    }
  }
  return 0;
}
