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>
#pragma warning(disable : 4996)
using namespace std;
const double pi = acos(-1);
const double eps = 1e-5;
const int INF = 0x3f3f3f3f;
const long long int INFLL = 0x3f3f3f3f3f3f3f3f;
const int maxn = 1e6 + 10;
int rev[maxn];
long long a[maxn];
int n, q;
struct STtree {
  struct tree {
    long long a;
    tree() : a(0) {}
  } node[maxn << 2];
  STtree() {}
  inline void push_up(int x, int d) {
    node[x].a = (node[(x << 1) ^ rev[d]].a + node[x << 1 | (1 ^ rev[d])].a);
  }
  void build(int x, int l, int r, long long* a, int d) {
    if (l == r) {
      node[x].a = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    build((x << 1) ^ rev[d], l, mid, a, d - 1);
    build(x << 1 | (1 ^ rev[d]), mid + 1, r, a, d - 1);
    push_up(x, d);
  }
  void update(int l, int r, int p, int x, long long k, int d) {
    if (l == r) {
      node[x].a = k;
      return;
    }
    int mid = (l + r) >> 1;
    if (p <= mid)
      update(l, mid, p, (x << 1) ^ rev[d], k, d - 1);
    else
      update(mid + 1, r, p, x << 1 | (1 ^ rev[d]), k, d - 1);
    push_up(x, d);
  }
  long long query(int l, int r, int left, int right, int x, int d) {
    long long res = 0;
    if (left <= l && r <= right) return node[x].a;
    int mid = (l + r) >> 1;
    if (left <= mid)
      res += query(l, mid, left, right, (x << 1) ^ rev[d], d - 1);
    if (right > mid)
      res += query(mid + 1, r, left, right, x << 1 | (1 ^ rev[d]), d - 1);
    return res;
  }
} st;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = (1); i <= (1 << n); i++) {
    cin >> a[i];
  }
  st.build(1, 1, 1 << n, a, n);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      long long x, k;
      cin >> x >> k;
      st.update(1, 1 << n, x, 1, k, n);
    } else if (op == 2) {
      int k;
      cin >> k;
      for (int i = (0); i <= (k); i++) rev[i] ^= 1;
    } else if (op == 3) {
      int k;
      cin >> k;
      rev[k + 1] ^= 1;
    } else {
      int l, r;
      cin >> l >> r;
      cout << st.query(1, 1 << n, l, r, 1, n) << '\n';
    }
  }
}
