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;
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
struct Node {
  long long sum = 0;
  Node *l = nullptr, *r = nullptr;
  bool rev[19];
  Node() {
    for (long long i = 0; i < 19; i++) rev[i] = false;
  }
};
vector<long long> a;
Node *build(long long l, long long r) {
  Node *it = new Node();
  if (l + 1 == r) {
    it->sum = a[l];
    return it;
  }
  long long m = (l + r) / 2;
  it->l = build(l, m);
  it->r = build(m, r);
  it->sum = (it->l->sum) + (it->r->sum);
  return it;
}
Node *upd(long long l, long long r, long long x, long long k, Node *it,
          long long d) {
  if (r <= x || x < l) return it;
  if (l + 1 == r) {
    it->sum = k;
    return it;
  }
  if ((it->rev)[d]) {
    swap(it->l, it->r);
    (it->rev)[d] = false;
    if (d) (it->rev)[d - 1] ^= 1;
  }
  for (long long i = 0; i < d; i++) {
    (it->l->rev)[i] ^= (it->rev)[i];
    (it->r->rev)[i] ^= (it->rev)[i];
    (it->rev)[i] = false;
  }
  long long m = (l + r) / 2;
  it->l = upd(l, m, x, k, it->l, d - 1);
  it->r = upd(m, r, x, k, it->r, d - 1);
  it->sum = (it->l->sum) + (it->r->sum);
  return it;
}
long long get(long long l, long long r, long long x, long long y, Node *it,
              long long d) {
  if (r <= x || y <= l) return 0;
  if (x <= l && r <= y) return it->sum;
  if ((it->rev)[d]) {
    swap(it->l, it->r);
    (it->rev)[d] = false;
    if (d) (it->rev)[d - 1] ^= 1;
  }
  for (long long i = 0; i < d; i++) {
    (it->l->rev)[i] ^= (it->rev)[i];
    (it->r->rev)[i] ^= (it->rev)[i];
    (it->rev)[i] = false;
  }
  long long m = (l + r) / 2;
  return get(l, m, x, y, it->l, d - 1) + get(m, r, x, y, it->r, d - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, q;
  cin >> n >> q;
  a.resize(1 << n);
  for (long long i = 0; i < (1 << n); i++) cin >> a[i];
  Node *it = build(0, (1 << n));
  for (long long i = 0; i < q; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      x--;
      it = upd(0, (1 << n), x, k, it, n);
    }
    if (t == 2) {
      long long k;
      cin >> k;
      (it->rev)[k] ^= 1;
    }
    if (t == 3) {
      long long k;
      cin >> k;
      if (k != n) {
        (it->rev)[k] ^= 1;
        (it->rev)[k + 1] ^= 1;
      }
    }
    if (t == 4) {
      long long l, r;
      cin >> l >> r;
      l--;
      cout << get(0, (1 << n), l, r, it, n) << '\n';
    }
  }
  return 0;
}
