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 GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << "{" << p.first << "," << p.second << "}";
  return os;
}
const int N = 3e5 + 5;
const int oo = 1e9 + 7;
struct segnode {
  int k;
  long long sum;
  int lazyreve;
  int rev;
  segnode *l, *r;
  segnode(long long x = 0) : k(0), sum(x), lazyreve(0), l(0), r(0), rev(0) {}
  segnode(segnode *l, segnode *r) : l(l), r(r) {
    sum = l->sum + r->sum;
    k = l->k + 1;
    lazyreve = 0;
    rev = 0;
  }
  void propagate() {
    if ((lazyreve >> k) & 1) {
      rev ^= 1;
      lazyreve ^= 1 << k;
    }
    if (rev) {
      swap(l, r);
      if (k) {
        l->rev ^= 1;
        r->rev ^= 1;
      }
      rev = 0;
    }
    if (k) {
      l->lazyreve ^= lazyreve;
      r->lazyreve ^= lazyreve;
    }
    lazyreve = 0;
  }
  void pr() {
    propagate();
    if (k == 0) {
      cout << sum << " ";
    } else {
      l->pr();
      r->pr();
    }
  }
  void repl(int ss, int se, int x, int k) {
    if (se < x or x < ss) return;
    propagate();
    if (ss == se) {
      sum = k;
    } else {
      int mid = (ss + se) / 2;
      l->repl(ss, mid, x, k);
      r->repl(mid + 1, se, x, k);
      sum = l->sum + r->sum;
    }
  }
  void revers(int k_) {
    lazyreve ^= 1 << k_;
    return;
  }
  long long sm(int ss, int se, int qs, int qe) {
    if (qe < ss or se < qs) return 0;
    propagate();
    if (qs <= ss and se <= qe) {
      return sum;
    } else {
      int mid = (ss + se) / 2;
      return l->sm(ss, mid, qs, qe) + r->sm(mid + 1, se, qs, qe);
    }
  }
};
int a[N];
segnode *build(int ss, int se) {
  if (ss == se) {
    return new segnode(a[ss]);
  } else {
    int mid = (ss + se) / 2;
    return new segnode(build(ss, mid), build(mid + 1, se));
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  int n2 = 1 << n;
  for (int i = 1; i <= n2; i++) {
    cin >> a[i];
  }
  segnode *root = build(1, n2);
  while (q--) {
    int t, x, k, l, r;
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      root->repl(1, n2, x, k);
    } else if (t == 2) {
      cin >> k;
      root->revers(k);
    } else if (t == 3) {
      cin >> k;
      root->revers(k + 1);
      root->revers(k);
    } else {
      cin >> l >> r;
      cout << root->sm(1, n2, l, r) << "\n";
    }
  }
}
