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;
template <typename T>
vector<T>& operator--(vector<T>& v) {
  for (auto& i : v) --i;
  return v;
}
template <typename T>
vector<T>& operator++(vector<T>& v) {
  for (auto& i : v) ++i;
  return v;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  for (auto& i : v) os << i << ' ';
  return os;
}
template <typename T, typename U>
istream& operator>>(istream& is, pair<T, U>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <typename T, typename U>
pair<T, U> operator-(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <typename T, typename U>
pair<T, U> operator+(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <typename T, typename U>
void umin(T& a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
void umax(T& a, U b) {
  if (a < b) a = b;
}
vector<long long> sums;
vector<bool> swapped;
vector<bool> reversed;
vector<long long> a;
int n;
void build(int i = 0, int l = 0, int r = n - 1) {
  if (l == r) {
    sums[i] = a[l];
    return;
  }
  int m = (l + r) / 2;
  build(i * 2 + 1, l, m);
  build(i * 2 + 2, m + 1, r);
  sums[i] = sums[i * 2 + 1] + sums[i * 2 + 2];
};
long long get_sum(int l, int r, int i = 0, int vl = 0, int vr = n - 1,
                  int lvl = 0) {
  l = max(l, vl);
  r = min(r, vr);
  if (r < l) return 0;
  42;
  ;
  if (l == vl && r == vr) {
    42;
    ;
    return sums[i];
  }
  int m = (vl + vr) / 2;
  int ln = (vr - vl + 1);
  if (!reversed[lvl] && !swapped[lvl]) {
    return get_sum(l, r, i * 2 + 1, vl, m, lvl + 1) +
           get_sum(l, r, i * 2 + 2, m + 1, vr, lvl + 1);
  } else if (!reversed[lvl] && swapped[lvl]) {
    return get_sum(l - ln / 2, r - ln / 2, i * 2 + 1, vl, m, lvl + 1) +
           get_sum(l + ln / 2, r + ln / 2, i * 2 + 2, m + 1, vr, lvl + 1);
  } else if (reversed[lvl] && !swapped[lvl]) {
    int l0 = (vr - r) + vl;
    int r0 = vr - (l - vl);
    return get_sum(l0, r0, i * 2 + 1, vl, m, lvl + 1) +
           get_sum(l0, r0, i * 2 + 2, m + 1, vr, lvl + 1);
  } else if (reversed[lvl] && swapped[lvl]) {
    int l0 = (vr - r) + vl;
    int r0 = vr - (l - vl);
    l = l0;
    r = r0;
    return get_sum(l - ln / 2, r - ln / 2, i * 2 + 1, vl, m, lvl + 1) +
           get_sum(l + ln / 2, r + ln / 2, i * 2 + 2, m + 1, vr, lvl + 1);
  }
};
void set_value(int ind, int k, int l = 0, int r = n - 1, int i = 0,
               int lvl = 0) {
  if (l == r) {
    sums[i] = k;
    return;
  }
  int m = (l + r) / 2;
  int ln = (r - l + 1);
  if (reversed[lvl]) {
    ind = r - (ind - l);
  }
  if (swapped[lvl]) {
    if (ind <= m) {
      ind += ln / 2;
    } else {
      ind -= ln / 2;
    }
  }
  if (ind <= m) {
    set_value(ind, k, l, m, i * 2 + 1, lvl + 1);
  } else {
    set_value(ind, k, m + 1, r, i * 2 + 2, lvl + 1);
  }
  sums[i] = sums[i * 2 + 1] + sums[i * 2 + 2];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int k, q;
  cin >> k >> q;
  n = (1 << k);
  a.resize(n);
  cin >> a;
  sums.assign(n * 2, 0);
  swapped.assign(k, false);
  reversed.assign(k + 1, false);
  build();
  while (q--) {
    int tp;
    cin >> tp;
    if (tp == 1) {
      int x, k;
      cin >> x >> k;
      set_value(x - 1, k);
    } else if (tp == 2) {
      int t;
      cin >> t;
      reversed[k - t] = !reversed[k - t];
    } else if (tp == 3) {
      int t;
      cin >> t;
      swapped[k - 1 - t] = !swapped[k - 1 - t];
    } else if (tp == 4) {
      int l, r;
      cin >> l >> r;
      --l;
      --r;
      cout << get_sum(l, r) << '\n';
    }
  }
  return 0;
}
