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;
int n, m, a[(1 << 19)], now = 0;
struct par {
  int x, len;
};
vector<par> q;
struct node {
  int l, r, mid;
  long long z;
  node *zuo, *you;
  node(int x, int y) : l(x), r(y), mid(l + r >> 1) {
    if (x < y) {
      zuo = new node(x, mid);
      you = new node(mid + 1, y);
      z = zuo->z + you->z;
    } else
      z = a[x], zuo = you = NULL;
  }
  void change(int x) {
    if (l == r) return (void)(z = a[x]);
    if (x <= mid)
      zuo->change(x);
    else
      you->change(x);
    z = zuo->z + you->z;
  }
  long long getsum(int x, int len) {
    if (r - l + 1 == len) return z;
    if (x <= mid)
      return zuo->getsum(x, len);
    else
      return you->getsum(x, len);
  }
  void split(int x, int y) {
    if (l == x && r == y) {
      q.push_back((par){x / (r - l + 1), r - l + 1});
      return;
    }
    if (y <= mid)
      zuo->split(x, y);
    else if (x >= mid + 1)
      you->split(x, y);
    else
      zuo->split(x, mid), you->split(mid + 1, y);
  }
} *root = NULL;
int main() {
  scanf("%d %d", &n, &m);
  n = 1 << n;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  root = new node(0, n - 1);
  for (int i = 1, id, x, y; i <= m; i++) {
    scanf("%d", &id);
    switch (id) {
      case 1:
        scanf("%d %d", &x, &y);
        x--;
        a[x ^ now] = y;
        root->change(x ^ now);
        break;
      case 2:
        scanf("%d", &x);
        now ^= (1 << x) - 1;
        break;
      case 3:
        scanf("%d", &x);
        now ^= 1 << x;
        break;
      case 4:
        scanf("%d %d", &x, &y);
        x--;
        y--;
        q.clear();
        root->split(x, y);
        long long ans = 0;
        for (int i = 0; i < q.size(); i++) {
          int c = q[i].len, d = 0;
          while (c % 2 == 0) c /= 2, d++;
          q[i].x ^= now >> d;
          ans += root->getsum(q[i].x * q[i].len, q[i].len);
        }
        printf("%lld\n", ans);
    }
  }
}
