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 <class P, class Q>
inline void smin(P &a, Q b) {
  if (b < a) a = b;
}
template <class P, class Q>
inline void smax(P &a, Q b) {
  if (a < b) a = b;
}
const int maxn = 18;
struct STE {
  long long sum;
  int rev, swp;
};
int n, q;
int a[1 << maxn];
STE seg[1 << (maxn + 1)];
void push(int x, int xlen) {
  int mask = (1 << xlen) - 1;
  seg[2 * x + 1].rev ^= seg[x].rev & mask;
  seg[2 * x + 1].swp ^= seg[x].swp & mask;
  seg[2 * x + 2].rev ^= seg[x].rev & mask;
  seg[2 * x + 2].swp ^= seg[x].swp & mask;
  seg[x].rev &= ~mask;
  seg[x].swp &= ~mask;
}
void pull(int x) { seg[x].sum = seg[2 * x + 1].sum + seg[2 * x + 2].sum; }
void modify(int x, int xl, int xlen, int qpos, int qval) {
  if (qpos < xl || xl + (1 << xlen) <= qpos) return;
  if (xlen == 0) {
    seg[x].sum = qval;
    return;
  }
  int xm = xl + (1 << (xlen - 1));
  if (seg[x].rev >> xlen & 1) qpos = 2 * xl + (1 << xlen) - qpos - 1;
  if (seg[x].swp >> xlen & 1) {
    if (qpos < xm)
      qpos += 1 << (xlen - 1);
    else
      qpos -= 1 << (xlen - 1);
  }
  push(x, xlen);
  modify(2 * x + 1, xl, xlen - 1, qpos, qval);
  modify(2 * x + 2, xm, xlen - 1, qpos, qval);
  pull(x);
}
long long getsum(int x, int xl, int xlen, int ql, int qr) {
  int xr = xl + (1 << xlen);
  if (qr <= xl || xr <= ql) return 0;
  if (ql <= xl && xr <= qr) return seg[x].sum;
  if (seg[x].rev >> xlen & 1) {
    ql = xl + xr - ql;
    qr = xl + xr - qr;
    swap(ql, qr);
  }
  int shift = 0;
  if (seg[x].swp >> xlen & 1) shift = 1 << (xlen - 1);
  int xm = xl + (1 << (xlen - 1));
  push(x, xlen);
  long long res = 0;
  res += getsum(2 * x + 1, xl, xlen - 1, ql - shift, qr - shift);
  res += getsum(2 * x + 2, xm, xlen - 1, ql + shift, qr + shift);
  return res;
}
void build(int x, int xl, int xlen) {
  if (xlen == 0) {
    seg[x].sum = a[xl];
    return;
  }
  int xm = xl + (1 << (xlen - 1));
  build(2 * x + 1, xl, xlen - 1);
  build(2 * x + 2, xm, xlen - 1);
  pull(x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0, i__n = (int)(1 << n); i < i__n; ++i) cin >> a[i];
  build(0, 0, n);
  for (int zz = 0, zz__n = (int)(q); zz < zz__n; ++zz) {
    int op;
    cin >> op;
    if (op == 1) {
      int pos, k;
      cin >> pos >> k;
      pos--;
      modify(0, 0, n, pos, k);
    } else if (op == 2 || op == 3) {
      int k;
      cin >> k;
      if (op == 2)
        seg[0].rev ^= 1 << k;
      else
        seg[0].swp ^= 1 << (k + 1);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      cout << getsum(0, 0, n, l, r) << endl;
    }
  }
  return 0;
}
