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 maxn = 3e5 + 9;
const int maxm = 6e5 + 9;
const long long inf = 1e18;
const int mod = 1e6;
int n, q;
long long tree[maxm];
int dep[maxm];
bool rev[50];
int read() {
  int a = 1, b = 0;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') a = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    b = (b << 3) + (b << 1) + c - '0';
    c = getchar();
  }
  return a * b;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
void up(int o) { tree[o] = tree[o << 1] + tree[o << 1 | 1]; }
void build(int o, int l, int r, int depth) {
  dep[o] = depth;
  if (l == r) {
    tree[o] = read();
    return;
  }
  int mid = (l + r) >> 1;
  build(o << 1, l, mid, depth - 1);
  build(o << 1 | 1, mid + 1, r, depth - 1);
  up(o);
}
void modify(int o, int l, int r, int pos, long long val) {
  if (l == r) {
    tree[o] = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    modify((o << 1) ^ rev[dep[o]], l, mid, pos, val);
  else
    modify((o << 1 | 1) ^ rev[dep[o]], mid + 1, r, pos, val);
  up(o);
}
long long query(int o, int l, int r, int L, int R) {
  if (l >= L && r <= R) {
    return tree[o];
  }
  int mid = (l + r) >> 1;
  long long sum = 0;
  if (L <= mid) sum += query((o << 1) ^ rev[dep[o]], l, mid, L, R);
  if (R > mid) sum += query((o << 1 | 1) ^ rev[dep[o]], mid + 1, r, L, R);
  return sum;
}
int main() {
  int opt, x, y;
  n = read(), q = read();
  for (register int i = 1; i <= n; ++i) rev[i] = 0;
  build(1, 1, 1 << n, n);
  n = (1 << n);
  while (q--) {
    opt = read();
    switch (opt) {
      case 1: {
        x = read(), y = read();
        modify(1, 1, n, x, y);
        break;
      }
      case 2: {
        x = read();
        for (register int i = 0; i <= x; ++i) rev[i] ^= 1;
        break;
      }
      case 3: {
        x = read();
        rev[x + 1] ^= 1;
        break;
      }
      case 4: {
        x = read(), y = read();
        print(query(1, 1, n, x, y));
        putchar('\n');
        break;
      }
    }
  }
  return 0;
}
