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>
std::mt19937_64 rng(
    std::chrono::steady_clock::now().time_since_epoch().count());
int myRand(int R) {
  int val = rng() % R;
  assert(val >= 0);
  return val;
}
const long double PI = atan(1.0) * 4;
const int64_t INF64 = 3e18;
const int32_t INF32 = 2e9 + 7;
const int32_t LOG = 21;
const int32_t MOD = 1e9 + 7;
using namespace std;
long long t[800005];
int N;
void build(vector<int> &a, int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = a[tl];
  } else {
    int tm = (tl + tr) / 2;
    build(a, v * 2, tl, tm);
    build(a, v * 2 + 1, tm + 1, tr);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
}
long long sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) {
    return t[v];
  }
  int tm = (tl + tr) / 2;
  return sum(v * 2, tl, tm, l, min(r, tm)) +
         sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
long long get_sum(int v, int tl, int tr, int l, int r, int x) {
  if (l > r) return 0;
  if (l == tl && r == tr) {
    int pow = r - l + 1;
    int l1 = (l ^ x) - (l ^ x) % pow;
    return sum(1, 0, N - 1, l1, l1 + pow - 1);
  }
  int tm = (tl + tr) / 2;
  return get_sum(v * 2, tl, tm, l, min(r, tm), x) +
         get_sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, x);
}
void update(int v, int tl, int tr, int pos, int new_val) {
  if (tl == tr) {
    t[v] = new_val;
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      update(v * 2, tl, tm, pos, new_val);
    else
      update(v * 2 + 1, tm + 1, tr, pos, new_val);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  N = 1 << n;
  vector<int> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  build(a, 1, 0, N - 1);
  int x = 0;
  for (int i = 0; i < q; i++) {
    int op;
    cin >> op;
    if (op == 1) {
      int pos, new_val;
      cin >> pos >> new_val;
      pos--;
      pos ^= x;
      update(1, 0, N - 1, pos, new_val);
    } else if (op == 2) {
      int k;
      cin >> k;
      x ^= ((1 << k) - 1);
    } else if (op == 3) {
      int k;
      cin >> k;
      x ^= (1 << k);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << get_sum(1, 0, N - 1, l, r, x) << '\n';
    }
  }
}
