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 N = (1 << 18), inf = 1e9;
int bit[18 + 2][N + 2], nn;
bool sg[18 + 2][8 * N + 16], lz[18 + 2][8 * N + 16];
void tooLazy(int id, int node, int lo, int hi) {
  if (!lz[id][node]) return;
  if (lo == hi)
    sg[id][node] ^= 1;
  else
    lz[id][node * 2] ^= 1, lz[id][node * 2 + 1] ^= 1;
  lz[id][node] = 0;
}
void upd(int id, int node, int lo, int hi, int lt, int rt) {
  tooLazy(id, node, lo, hi);
  if (lo > rt || hi < lt) return;
  if (lo >= lt && hi <= rt) {
    lz[id][node] ^= 1;
    tooLazy(id, node, lo, hi);
    return;
  }
  int md = (lo + hi) / 2;
  upd(id, node * 2, lo, md, lt, rt);
  upd(id, node * 2 + 1, md + 1, hi, lt, rt);
}
int get(int id, int node, int lo, int hi, int ind) {
  tooLazy(id, node, lo, hi);
  if (lo == hi) return sg[id][node];
  int md = (lo + hi) / 2;
  if (ind <= md)
    return get(id, node * 2, lo, md, ind);
  else
    return get(id, node * 2 + 1, md + 1, hi, ind);
}
int sz = 1, a[N + 2];
struct segment {
  int l, r, vl;
  long long sum;
} seg[2 * N + 8];
void build(int node, int lo, int hi) {
  for (int i = lo; i <= hi; i++) seg[node].sum += a[i];
  if (lo == hi) {
    seg[node].vl = a[lo];
    return;
  }
  seg[node].l = ++sz;
  seg[node].r = ++sz;
  int md = (lo + hi) / 2;
  build(seg[node].l, lo, md);
  build(seg[node].r, md + 1, hi);
}
int delta;
void rep(int l, int node, int lo, int hi, int id, int vl) {
  if (get(l, 1, 1, 2 * nn, node)) {
    swap(seg[node].l, seg[node].r);
    upd(l, 1, 1, 2 * nn, node, node);
  }
  if (lo == hi) {
    delta = vl - seg[node].vl;
    seg[node].vl += delta;
    seg[node].sum += delta;
    return;
  }
  int md = (lo + hi) / 2;
  if (id <= md)
    rep(l + 1, seg[node].l, lo, md, id, vl);
  else
    rep(l + 1, seg[node].r, md + 1, hi, id, vl);
  seg[node].sum += delta;
}
long long sm(int l, int node, int lo, int hi, int lt, int rt) {
  if (get(l, 1, 1, 2 * nn, node)) {
    swap(seg[node].l, seg[node].r);
    upd(l, 1, 1, 2 * nn, node, node);
  }
  if (lo > rt || hi < lt) return 0;
  if (lo >= lt && hi <= rt) return seg[node].sum;
  int md = (lo + hi) / 2;
  return sm(l + 1, seg[node].l, lo, md, lt, rt) +
         sm(l + 1, seg[node].r, md + 1, hi, lt, rt);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  nn = (1 << n);
  for (int i = 1; i <= nn; i++) cin >> a[i];
  build(1, 1, nn);
  while (q--) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int x, k;
      cin >> x >> k;
      rep(0, 1, 1, nn, x, k);
    } else if (ty == 2) {
      int k;
      cin >> k;
      for (int i = n - k; i < n; i++) upd(i, 1, 1, 2 * nn, 1, 2 * nn);
    } else if (ty == 3) {
      int k;
      cin >> k;
      upd(n - k - 1, 1, 1, 2 * nn, 1, 2 * nn);
    } else {
      int l, r;
      cin >> l >> r;
      cout << sm(0, 1, 1, nn, l, r) << '\n';
    }
  }
  return 0;
}
