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\}. 
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

n,q = map(int,input().split())

SegmentTree = []

for i in range(n + 1):
    SegmentTree.append([0] * (1 << (n - i)))

a = list(map(int,input().split()))

for i in range(1 << n):
    for j in range(n + 1):
        SegmentTree[j][i >> j] += a[i]

QueryList = [0] * n

def FindSum(m): 
    summ = 0
    for i in range(n + 1):
        if m & (1 << i):
            m -= 1 << i
            mCpy = m
            for j in range(i,n):
                if QueryList[j]:
                    mCpy ^= ((1 << (j+1)) - (1 << i))
            summ += SegmentTree[i][mCpy >> i]
    return summ

for _ in range(q):
    t = list(map(int,input().split()))
    if t[0] == 1:
        loc = t[1] - 1
        for i in range(n):
            if QueryList[i]:loc ^= ((1 << (i + 1)) - 1)
        increment = t[2] - SegmentTree[0][loc]
        for j in range(n + 1):SegmentTree[j][loc >> j] += increment
    if t[0] == 2:
        if t[1] == 0:continue
        QueryList[t[1] - 1] = 1 - QueryList[t[1] - 1]
    if t[0] == 3:
        QueryList[t[1]] = 1 - QueryList[t[1]]
        if t[1] == 0:continue
        QueryList[t[1] - 1] = 1 - QueryList[t[1] - 1]    
    if t[0] == 4:print(FindSum(t[2]) - FindSum(t[1] - 1))    