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>
const int inf = 0x3f3f3f3f;
using namespace std;
template <typename T = int>
T yh() {
  T ret = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == EOF) return -1;
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) ret = (ret << 3) + (ret << 1) + (c ^ 48), c = getchar();
  return f ? -ret : ret;
}
template <typename T>
void yh(T& ret) {
  ret = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == EOF) return ret = -1, void();
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) ret = (ret << 3) + (ret << 1) + (c ^ 48), c = getchar();
  ret = f ? -ret : ret;
}
const int maxn = 3e5 + 5;
int n, m;
long long sum[maxn << 2];
int cs[maxn << 2];
bool rev[100];
long long ans = 0;
void push_up(int v) { sum[v] = sum[(v << 1)] + sum[(v << 1 | 1)]; }
void build(int v, int l, int r, int cs) {
  ::cs[v] = cs;
  if (l == r) return sum[v] = yh(), void();
  build((v << 1), l, ((l + r) >> 1), cs - 1);
  build((v << 1 | 1), ((l + r) >> 1) + 1, r, cs - 1);
  push_up(v);
}
void modify(int v, int l, int r, int p, int val) {
  if (l == r) {
    sum[v] = val;
    return;
  }
  if (rev[cs[v]]) {
    if (p <= ((l + r) >> 1))
      modify((v << 1 | 1), l, ((l + r) >> 1), p, val);
    else
      modify((v << 1), ((l + r) >> 1) + 1, r, p, val);
  } else {
    if (p <= ((l + r) >> 1))
      modify((v << 1), l, ((l + r) >> 1), p, val);
    else
      modify((v << 1 | 1), ((l + r) >> 1) + 1, r, p, val);
  }
  push_up(v);
}
void ask(int v, int l, int r, int al, int ar) {
  if (al <= l && ar >= r) {
    ans += sum[v];
    return;
  }
  if (rev[cs[v]]) {
    if (al <= ((l + r) >> 1)) ask((v << 1 | 1), l, ((l + r) >> 1), al, ar);
    if (ar > ((l + r) >> 1)) ask((v << 1), ((l + r) >> 1) + 1, r, al, ar);
  } else {
    if (al <= ((l + r) >> 1)) ask((v << 1), l, ((l + r) >> 1), al, ar);
    if (ar > ((l + r) >> 1)) ask((v << 1 | 1), ((l + r) >> 1) + 1, r, al, ar);
  }
  push_up(v);
}
int main() {
  n = yh();
  m = yh();
  build(1, 1, (1 << n), n);
  n = (1 << n);
  while (m--) {
    int opt = yh(), x = yh();
    switch (opt) {
      case 1: {
        modify(1, 1, n, x, yh());
        break;
      }
      case 2: {
        for (register int k = (0); k <= (x); k++) rev[k] ^= 1;
        break;
      }
      case 3: {
        rev[x + 1] ^= 1;
        break;
      }
      case 4: {
        ans = 0;
        ask(1, 1, n, x, yh());
        cout << ans << '\n';
      }
    }
  }
  return 0;
}
