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 = 2e6 + 10;
const int inf = 1e9 + 7;
const long long INF = 1e18 + 7;
struct Tree {
  Tree *l, *r;
  int fl;
  long long s;
  Tree() {
    l = r = NULL;
    fl = 0;
  }
};
typedef Tree *tree;
tree t;
int n, q;
void build(tree t, int L, int R) {
  if (L == R) {
    cin >> t->s;
  } else {
    int m = (L + R) / 2;
    tree l = new Tree(), r = new Tree();
    t->l = l, t->r = r;
    build(t->l, L, m);
    build(t->r, m + 1, R);
    t->s = t->l->s + t->r->s;
  }
}
void push(tree &t, int L, int R) {
  if ((t->fl & (R - L + 1)) && R > L) {
    swap(t->l, t->r);
  }
  if (R > L) {
    t->l->fl ^= t->fl;
    t->r->fl ^= t->fl;
    t->fl = 0;
  }
}
void upd(tree &t, int L, int R, int i, int x) {
  push(t, L, R);
  if (L == R) {
    t->s = x;
  } else {
    int m = (L + R) / 2;
    if (i <= m)
      upd(t->l, L, m, i, x);
    else
      upd(t->r, m + 1, R, i, x);
    t->s = t->l->s + t->r->s;
  }
}
long long get(tree &t, int L, int R, int l, int r) {
  push(t, L, R);
  if (R < l || L > r || L > R) return 0;
  if (l <= L && R <= r) {
    return t->s;
  } else {
    int m = (L + R) / 2;
    return get(t->l, L, m, l, r) + get(t->r, m + 1, R, l, r);
  }
}
void print(tree &t, int L, int R) {
  push(t, L, R);
  if (L == R) {
    cout << t->s << " ";
  } else {
    int m = (L + R) >> 1;
    print(t->l, L, m);
    print(t->r, m + 1, R);
  }
}
int tp, l, r;
int main() {
  cin >> n >> q;
  t = new Tree();
  build(t, 1, (1 << n));
  while (q--) {
    cin >> tp >> l;
    if (tp == 1) {
      cin >> r;
      upd(t, 1, (1 << n), l, r);
    }
    if (tp == 2) t->fl ^= ((1 << (l + 1)) - 1);
    if (tp == 3) t->fl ^= (1 << (l + 1));
    if (tp == 4) {
      cin >> r;
      cout << get(t, 1, (1 << n), l, r) << endl;
    }
  }
}
