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 unsigned long long MOD = 119 << 23 | 1;
unsigned long long gcd(unsigned long long x, unsigned long long y) {
  if (!y) return x;
  return gcd(y, x % y);
}
unsigned long long inv(unsigned long long x) {
  unsigned long long r = 1ll, t = x, k = MOD - 2ll;
  while (k > 0) {
    if (k & 1ll) r = (long long)r * t % MOD;
    t = (long long)t * t % MOD;
    k >>= 1;
  }
  return r;
}
void max_self(int &_a, int _b) { _a = max(_a, _b); }
const int nax = 2e5 + 13;
int n, k, x, q, pocz, kon;
int leafcount = 1;
vector<unsigned long long> T;
bool swaped[20], reversed[20];
void readIN() {
  cin >> n >> q;
  T.resize((2 << n) + 13);
  for (int i = 0; i < (1 << n); i++) cin >> T[(1 << n) + i];
  for (int i = (1 << n) - 1; i > 0; i--) T[i] = T[2 * i] + T[2 * i + 1];
}
unsigned long long query(int akt, int l, int r, int L, int R, int lvl) {
  if (reversed[lvl]) {
    int r1 = l - L;
    int r2 = R - r;
    r = R - r1;
    l = L + r2;
  }
  if (swaped[lvl]) {
    if (akt % 2 == 0) {
      akt++;
      L += (1 << lvl);
      R += (1 << lvl);
      l += (1 << lvl);
      r += (1 << lvl);
    } else {
      akt--;
      L -= (1 << lvl);
      R -= (1 << lvl);
      l -= (1 << lvl);
      r -= (1 << lvl);
    }
  }
  if (l == L && r == R) return T[akt];
  if (l > r) return 0;
  int mid = (L + R) / 2;
  return query(akt * 2, l, min(mid, r), L, mid, lvl - 1) +
         query(akt * 2 + 1, max(mid + 1, l), r, mid + 1, R, lvl - 1);
}
void replace(int pos, int val) {
  int lvl = n, akt = 1, L = 1, R = (1 << n);
  while (akt < (1 << n)) {
    if (swaped[lvl]) {
      if (akt % 2 == 0) {
        akt++;
        L += (1 << lvl);
        R += (1 << lvl);
        pos += (1 << lvl);
      } else {
        akt--;
        L -= (1 << lvl);
        R -= (1 << lvl);
        pos -= (1 << lvl);
      }
    }
    if (reversed[lvl]) {
      int r = pos - L;
      pos = R - r;
    }
    int mid = (L + R) / 2;
    if (pos <= mid) {
      akt = akt * 2;
      R = mid;
    } else {
      akt = akt * 2 + 1;
      L = mid + 1;
    }
    lvl--;
  }
  if (swaped[lvl]) {
    if (akt % 2 == 0) {
      akt++;
      L += (1 << lvl);
      R += (1 << lvl);
      pos += (1 << lvl);
    } else {
      akt--;
      L -= (1 << lvl);
      R -= (1 << lvl);
      pos -= (1 << lvl);
    }
  }
  T[akt] = val;
  akt /= 2;
  while (akt > 0) {
    T[akt] = T[akt * 2] + T[akt * 2 + 1];
    akt /= 2;
  }
}
void solve() {
  while (q--) {
    int a;
    cin >> a;
    if (a == 1) {
      cin >> x >> k;
      replace(x, k);
    } else if (a == 2) {
      cin >> k;
      reversed[k] = !reversed[k];
    } else if (a == 3) {
      cin >> k;
      swaped[k] = !swaped[k];
    } else {
      cin >> pocz >> kon;
      cout << query(1, pocz, kon, 1, (1 << n), n) << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  int t = 1;
  while (t--) {
    readIN();
    solve();
  }
  return 0;
}
