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 LOG = 19;
const int MAXN = 1 << LOG;
int n, q;
int N;
long long a[MAXN];
struct Node {
  Node *l, *r;
  long long sum;
  int rev[LOG];
  int swaps[LOG];
  Node() {
    l = NULL;
    r = NULL;
    for (int i = 0; i <= n; ++i) rev[i] = swaps[i] = 0;
  }
  void init(long long val) { sum = val; }
  void merge() { sum = l->sum + r->sum; }
  void push(int h, int i, int j) {
    int mid = (i + j) >> 1;
    if (!h) return;
    for (int k = 0; k <= n; ++k) {
      l->rev[k] = (l->rev[k] + rev[k]) % 2;
      r->rev[k] = (r->rev[k] + rev[k]) % 2;
      l->swaps[k] = (l->swaps[k] + swaps[k]) % 2;
      r->swaps[k] = (r->swaps[k] + swaps[k]) % 2;
    }
    if (rev[h]) {
      swap(l, r);
      l->rev[h - 1] = (l->rev[h - 1] + rev[h]) % 2;
      r->rev[h - 1] = (r->rev[h - 1] + rev[h]) % 2;
    }
    if (swaps[h]) swap(l, r);
    for (int k = 0; k <= n; ++k) swaps[k] = rev[k] = 0;
  }
};
Node *build(Node *root, int i, int j) {
  root = new Node();
  if (i == j) {
    root->init(a[i]);
    return root;
  }
  int mid = (i + j) >> 1;
  root->l = build(root->l, i, mid);
  root->r = build(root->r, mid + 1, j);
  root->merge();
  return root;
}
void update(Node *root, int h, int i, int j, int pos, long long val) {
  root->push(h, i, j);
  if (i == j) {
    root->sum = val;
    return;
  }
  int mid = (i + j) >> 1;
  if (pos <= mid)
    update(root->l, h - 1, i, mid, pos, val);
  else
    update(root->r, h - 1, mid + 1, j, pos, val);
  root->merge();
}
long long query(Node *root, int h, int i, int j, int l, int r) {
  root->push(h, i, j);
  if (i > j || j < l || r < i) return 0LL;
  if (l <= i && j <= r) return root->sum;
  int mid = (i + j) >> 1;
  return query(root->l, h - 1, i, mid, l, r) +
         query(root->r, h - 1, mid + 1, j, l, r);
}
Node *seg;
int main() {
  scanf("%d %d", &n, &q);
  N = (1 << n);
  for (int i = 1; i <= N; ++i) scanf("%lld", &a[i]);
  seg = build(seg, 1, N);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int x, k;
      scanf("%d %d", &x, &k);
      update(seg, n, 1, N, x, k);
    } else if (type == 4) {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%lld\n", query(seg, n, 1, N, l, r));
    } else if (type == 2) {
      int k;
      scanf("%d", &k);
      if (k) {
        seg->rev[k] = (seg->rev[k] + 1) % 2;
        seg->push(n, 1, N);
      }
    } else {
      int k;
      scanf("%d", &k);
      seg->swaps[k + 1] = (seg->swaps[k + 1] + 1) % 2;
      seg->push(n, 1, N);
    }
  }
  return 0;
}
