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 long long INF = ~(1 << 31);
const double pi = acos(-1);
const double EPS = 1e-9;
template <class T>
T smod(T a, T b) {
  return (a % b + b) % b;
}
struct node {
  long long val;
  node(long long _val) { val = _val; };
};
long long XOR = 0;
struct segment_tree {
  long long sz, n;
  vector<node> t;
  segment_tree(long long _n) : sz(_n * 4), n(_n) {
    for (long long i = (0); i < (sz); i++) t.push_back(node(0));
  }
  void build(vector<long long> &a, long long v = 1, long long tl = 0,
             long long tr = -1) {
    if (tr == -1) tr = n - 1;
    if (tl == tr) {
      t[v].val = a[tl];
    } else {
      long long tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = node(t[v * 2].val + t[v * 2 + 1].val);
    }
  }
  node query(long long l, long long r, long long height, long long v = 1,
             long long tl = 0, long long tr = -1) {
    if (tr == -1) tr = n - 1;
    if (l > r) return node(0);
    if (l == tl && r == tr) return t[v];
    long long tm = (tl + tr) / 2;
    if (XOR & (1 << height)) {
      return node(query(l, min(r, tm), height - 1, v * 2 + 1, tl, tm).val +
                  query(max(l, tm + 1), r, height - 1, v * 2, tm + 1, tr).val);
    }
    return node(
        query(l, min(r, tm), height - 1, v * 2, tl, tm).val +
        query(max(l, tm + 1), r, height - 1, v * 2 + 1, tm + 1, tr).val);
  }
  void update(long long pos, long long new_val, long long height,
              long long v = 1, long long tl = 0, long long tr = -1) {
    if (tr == -1) tr = n - 1;
    if (tl == tr)
      t[v].val = new_val;
    else {
      long long tm = (tl + tr) / 2;
      if (XOR & (1 << height)) {
        if (pos <= tm)
          update(pos, new_val, height - 1, v * 2 + 1, tl, tm);
        else
          update(pos, new_val, height - 1, v * 2, tm + 1, tr);
      } else {
        if (pos <= tm)
          update(pos, new_val, height - 1, v * 2, tl, tm);
        else
          update(pos, new_val, height - 1, v * 2 + 1, tm + 1, tr);
      }
      t[v].val = t[v * 2].val + t[v * 2 + 1].val;
    }
  }
};
int main() {
  cin.sync_with_stdio(false);
  long long n, q;
  cin >> n >> q;
  segment_tree sq(1 << n);
  vector<long long> A(1 << n);
  for (long long i = (0); i < (1 << n); i++) cin >> A[i];
  sq.build(A);
  for (long long i = (0); i < (q); i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      x--;
      sq.update(x, k, n);
    } else if (t == 4) {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      cout << sq.query(l, r, n).val << endl;
    } else if (t == 3) {
      long long k;
      cin >> k;
      XOR ^= (1 << (k + 1));
    } else if (t == 2) {
      long long k;
      cin >> k;
      XOR ^= (1 << (k + 1)) - 1;
    }
  }
  return 0;
}
