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;
inline int read(void) {
  register int x = 0, f = 1, ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 300005;
struct node {
  long long sum;
  int tag;
  node *left, *right;
} seg[maxn << 1], *root = seg;
int tot_seg, a[maxn];
int Log[maxn];
void build(int l, int r, node *p = root) {
  p->tag = 0;
  if (l == r) {
    p->sum = a[l];
    return;
  }
  register int mid = (l + r) >> 1;
  p->left = seg + ++tot_seg;
  build(l, mid, p->left);
  p->right = seg + ++tot_seg;
  build(mid + 1, r, p->right);
  p->sum = p->left->sum + p->right->sum;
}
inline void pushdown(int l, int r, node *p) {
  if (p->tag) {
    p->left->tag ^= p->tag;
    p->right->tag ^= p->tag;
    p->tag = 0;
  }
}
inline void renew(int l, int r, node *p) {
  if (l == r) return;
  register int c = Log[r - l + 1] - 1;
  if (p->tag & (1 << c)) {
    swap(p->left, p->right);
    p->tag ^= (1 << c);
    p->tag ^= (1 << (c - 1));
  }
}
void modify(int l, int r, int c, int d, node *p = root) {
  renew(l, r, p);
  if (l == r) {
    p->sum = d;
    return;
  }
  pushdown(l, r, p);
  register int mid = (l + r) >> 1;
  if (c <= mid)
    modify(l, mid, c, d, p->left);
  else
    modify(mid + 1, r, c, d, p->right);
  p->sum = p->left->sum + p->right->sum;
}
long long query(int l, int r, int ql, int qr, node *p = root) {
  renew(l, r, p);
  if (ql <= l && r <= qr) return p->sum;
  pushdown(l, r, p);
  register int mid = (l + r) >> 1;
  register long long res = 0;
  if (ql <= mid) res += query(l, mid, ql, qr, p->left);
  if (mid < qr) res += query(mid + 1, r, ql, qr, p->right);
  return res;
}
signed main(void) {
  int n = read(), m = read();
  for (register int i = 0; i <= n; ++i) Log[1 << i] = i;
  n = 1 << n;
  for (register int i = 1; i <= n; ++i) a[i] = read();
  build(1, n);
  while (m--) {
    int opt = read();
    if (opt == 1) {
      int x = read(), y = read();
      modify(1, n, x, y);
    } else if (opt == 2) {
      int x = read();
      if (x > 0) root->tag ^= (1 << (x - 1));
    } else if (opt == 3) {
      int x = read();
      root->tag ^= (1 << x);
      if (x > 0) root->tag ^= (1 << (x - 1));
    } else if (opt == 4) {
      int x = read(), y = read();
      printf("%lld\n", query(1, n, x, y));
    }
  }
  return 0;
}
