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;
vector<long long> sieve;
void Sieve(int N) {
  const long long maxn = N;
  sieve.resize(maxn);
  for (long long(i) = 0; (i) < (maxn); ++(i)) sieve[i] = i;
  sieve[1] = -1;
  sieve[0] = -1;
  for (long long(i) = 2; (i) <= (maxn); (i) += (1))
    if (i == sieve[i])
      for (long long j = 2 * i; j < maxn; j += i)
        if (sieve[j] == j) sieve[j] = i;
}
long long extended_GCD(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a, long long mod = 1000000007) {
  long long x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
long long n;
long long const N = 2 * 1e6 + 2;
long long a[N], seg[4 * (1 << 18)], ht;
long long level[20];
void build(long long low, long long high, long long pos) {
  if (low == high) {
    seg[pos] = a[low];
    return;
  }
  long long mid = (low + high) >> 1;
  build(low, mid, 2 * pos + 1);
  build(mid + 1, high, 2 * pos + 2);
  seg[pos] = seg[2 * pos + 1] + seg[2 * pos + 2];
}
void update(long long upos, long long val, long long low, long long high,
            long long pos, long long lev) {
  if (low == high) {
    seg[pos] = a[upos] = val;
    return;
  }
  if (level[lev]) {
    long long mid = (low + high) >> 1;
    if (upos <= mid) {
      update(upos, val, low, mid, pos * 2 + 2, lev - 1);
    } else {
      update(upos, val, mid + 1, high, pos * 2 + 1, lev - 1);
    }
    seg[pos] = seg[2 * pos + 1] + seg[2 * pos + 2];
  } else {
    long long mid = (low + high) >> 1;
    if (upos <= mid) {
      update(upos, val, low, mid, pos * 2 + 1, lev - 1);
    } else {
      update(upos, val, mid + 1, high, pos * 2 + 2, lev - 1);
    }
    seg[pos] = seg[2 * pos + 1] + seg[2 * pos + 2];
  }
}
long long query(long long low, long long high, long long l, long long r,
                long long pos, long long lev) {
  if (low > r || l > high) return 0;
  if (l <= low and r >= high) {
    return seg[pos];
  }
  if (level[lev]) {
    long long mid = (low + high) >> 1;
    return query(mid + 1, high, l, r, 2 * pos + 1, lev - 1) +
           query(low, mid, l, r, 2 * pos + 2, lev - 1);
  } else {
    long long mid = (low + high) >> 1;
    return query(low, mid, l, r, 2 * pos + 1, lev - 1) +
           query(mid + 1, high, l, r, 2 * pos + 2, lev - 1);
  }
}
void solve() {
  cin >> n;
  long long N = n;
  n = 1 << n;
  long long q;
  cin >> q;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  build(0, n - 1, 0);
  for (int zz = 0; zz < q; ++zz) {
    long long c;
    cin >> c;
    if (c == 1) {
      long long x, k;
      cin >> x >> k;
      x--;
      update(x, k, 0, n - 1, 0, N);
    }
    if (c == 3) {
      long long k;
      cin >> k;
      level[k + 1] ^= 1;
    }
    if (c == 2) {
      long long k;
      cin >> k;
      for (int i = 1; i <= k; ++i) {
        level[i] ^= 1;
      }
    }
    if (c == 4) {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      long long ans = query(0, n - 1, l, r, 0, N);
      cout << ans << "\n";
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
