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;
struct st {
  int n;
  vector<int> t;
  st(int sz) {
    n = sz;
    t = vector<int>(2 * n);
  }
  void upd(int l, int r, int val) {
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) t[l++] += val;
      if (r & 1) t[--r] += val;
    }
  }
  int qry(int pos) {
    int ans = 0;
    for (pos += n; pos; pos >>= 1) ans += t[pos];
    return ans;
  }
};
int n, m;
vector<vector<long long>> arr;
vector<vector<int>> lef, rig, par;
vector<st> rev, swp;
void init() {
  for (int j = 0; j < m; j++) par[0][j] = j >> 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < arr[i].size(); j++) {
      arr[i][j] = arr[i - 1][j << 1] + arr[i - 1][j << 1 | 1];
      lef[i][j] = j << 1;
      rig[i][j] = j << 1 | 1;
      par[i][j] = j >> 1;
    }
  }
}
void build(int x) {
  stack<int> sta;
  int pos = 0;
  for (int i = n; i > 0; i--) {
    sta.push(pos);
    int tmp = i - 1;
    if (x & (1 << tmp))
      pos = rig[i][pos];
    else
      pos = lef[i][pos];
  }
  for (int i = 1; i <= n; i++) {
    pos = sta.top();
    sta.pop();
    arr[i][pos] = arr[i - 1][lef[i][pos]] + arr[i - 1][rig[i][pos]];
  }
}
int push(int x) {
  int pos = 0;
  for (int i = n; i > 0; i--) {
    int swp_cur = swp[i].qry(pos);
    int rev_cur = rev[i].qry(pos);
    int move = (swp_cur + rev_cur) % 2;
    swp[i].upd(pos, pos + 1, -swp_cur);
    rev[i].upd(pos, pos + 1, -rev_cur);
    int left = lef[i][pos], right = rig[i][pos];
    rev[i - 1].upd(left, left + 1, rev_cur % 2);
    rev[i - 1].upd(right, right + 1, rev_cur % 2);
    if (move) {
      swap(lef[i][pos], rig[i][pos]);
      swap(left, right);
    }
    int tmp = i - 1;
    if (x & (1 << tmp))
      pos = right;
    else
      pos = left;
  }
  return pos;
}
void replace(int x, int k) {
  int pos = push(x);
  arr[0][pos] = k;
  build(x);
}
void reverse(int k) { rev[k].upd(0, rev[k].n, 1); }
void swap(int k) {
  ++k;
  swp[k].upd(0, swp[k].n, 1);
}
long long sum(int i, int pos, int left, int right, int l, int r) {
  if (left == l && right == r) return arr[i][pos];
  int mid = (left + right) / 2;
  long long res = 0;
  if (l <= mid) res += sum(i - 1, lef[i][pos], left, mid, l, min(mid, r));
  if (mid < r)
    res += sum(i - 1, rig[i][pos], mid + 1, right, max(mid + 1, l), r);
  return res;
}
long long sum(int l, int r) {
  push(l);
  push(r);
  return sum(n, 0, 0, m - 1, l, r);
}
void debug() {
  return;
  for (int i = 0; i < m; i++) push(i);
  queue<int> qq;
  qq.push(0);
  for (int i = n; i > 0; i--) {
    int sz = qq.size();
    while (sz--) {
      int cur = qq.front();
      qq.pop();
      cout << arr[i][cur] << ' ';
      qq.push(lef[i][cur]);
      qq.push(rig[i][cur]);
    }
    cout << '\n';
  }
  while (!qq.empty()) {
    cout << arr[0][qq.front()] << ' ';
    qq.pop();
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> n >> q;
  m = 1 << n;
  for (int i = n; i >= 0; i--) {
    arr.push_back(vector<long long>(1 << i));
    lef.push_back(vector<int>(1 << i));
    rig.push_back(vector<int>(1 << i));
    par.push_back(vector<int>(1 << i));
    rev.push_back(st(1 << i));
    swp.push_back(st(1 << i));
  }
  for (int i = 0; i < m; i++) cin >> arr[0][i];
  init();
  while (q--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int x, k;
      cin >> x >> k;
      replace(--x, k);
    } else if (tp == 2) {
      int k;
      cin >> k;
      reverse(k);
    } else if (tp == 3) {
      int k;
      cin >> k;
      swap(k);
    } else {
      int l, r;
      cin >> l >> r;
      --l;
      --r;
      cout << sum(l, r) << '\n';
      debug();
    }
  }
}
