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;
void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
int ctz(unsigned x) { return __builtin_ctz(x); }
int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
int clz(unsigned x) { return __builtin_clz(x); }
int clzll(unsigned long long x) { return __builtin_clzll(x); }
int popcnt(unsigned x) { return __builtin_popcount(x); }
int popcntll(unsigned long long x) { return __builtin_popcountll(x); }
int bsr(unsigned x) { return 31 ^ clz(x); }
int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
int mask;
int a[(1 << 18)];
long long tree[2 * (1 << 18)];
void update(int u, int b, int sz, int x, int k) {
  if (sz == 1) {
    tree[u] = k;
    return;
  }
  sz /= 2;
  int lc = 2 * u, rc = 2 * u + 1;
  if (x < b + sz)
    update(lc, b, sz, x, k);
  else
    update(rc, b + sz, sz, x, k);
  tree[u] = tree[lc] + tree[rc];
}
long long query(int u, int b, int sz, int l, int r) {
  int e = b + sz;
  if (b >= r || l >= e) return 0;
  if (b >= l && r >= e) return tree[u];
  sz /= 2;
  int lc = 2 * u, rc = 2 * u + 1;
  if (mask & sz) swap(lc, rc);
  return query(lc, b, sz, l, r) + query(rc, b + sz, sz, l, r);
}
int main() {
  canhazfast();
  int n, m, q;
  cin >> n >> q;
  m = 1 << n;
  for (int i = 0; i < m; ++i) cin >> a[i];
  for (int i = m - 1; i >= 0; --i) tree[m + i] = a[i];
  for (int i = m - 1; i > 0; --i) tree[i] = tree[2 * i] + tree[2 * i + 1];
  for (int i = 0; i < q; ++i) {
    int t, k, x, l, r;
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      x = (x - 1) ^ mask;
      update(1, 0, m, x, k);
      continue;
    }
    if (t == 4) {
      cin >> l >> r;
      cout << query(1, 0, m, l - 1, r) << '\n';
      continue;
    }
    cin >> k;
    x = 1 << k;
    if (t % 2 == 0) --x;
    mask ^= x;
  }
  return 0;
}
