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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
int cansel_sync = (ios::sync_with_stdio(0), cin.tie(0), 0);
const int N = 1000010;
const int inf = ~0u >> 2;
const long long INF = ~0ull >> 2;
long long read() {
  long long x;
  if (scanf("%lld", &x) == -1) exit(0);
  return x;
}
const double pi = acos(-1.0);
double readf() {
  double x;
  if (scanf("%lf", &x) == -1) exit(0);
  return x;
}
template <typename T>
void operator<<(vector<T> &a, T b) {
  a.push_back(b);
}
const long long mod = (1 ? 1000000007 : 998244353);
long long mul(long long a, long long b, long long m = mod) { return a * b % m; }
long long qpow(long long a, long long b, long long m = mod) {
  long long ans = 1;
  for (; b; a = mul(a, a, m), b >>= 1)
    if (b & 1) ans = mul(ans, a, m);
  return ans;
}
long long in[N];
struct seg {
  void toz(long long x1, long long x2) { z1 ^= x1, z2 ^= x2; }
  long long a, z1, z2;
  seg *lc, *rc;
  void init(long long, long long);
  void up() { a = (lc->a + rc->a); }
  void down(long long l, long long r) {
    if (z1 & 1) swap(lc, rc), z1 ^= 3;
    if (z2 & 1) swap(lc, rc);
    if (l < r) {
      lc->toz(z1 >> 1, z2 >> 1);
      rc->toz(z1 >> 1, z2 >> 1);
    }
    z1 = z2 = 0;
  }
  void update(long long l, long long r, long long k1, long long k2) {
    z1 ^= k1;
    z2 ^= k2;
    down(l, r);
  }
  void change(long long l, long long r, long long x, long long y) {
    if (l == x && r == x) {
      down(l, r);
      a = y;
      return;
    }
    if (x < l || x > r) {
      down(l, r);
      return;
    }
    long long m = (l + r) / 2;
    down(l, r);
    lc->change(l, m, x, y);
    rc->change(m + 1, r, x, y);
    up();
  }
  long long query(long long l, long long r, long long x, long long y) {
    x = max(x, l);
    y = min(y, r);
    if (x > y) return 0;
    down(l, r);
    if (x == l && y == r) return a;
    long long m = (l + r) / 2;
    return (lc->query(l, m, x, y) + rc->query(m + 1, r, x, y));
  }
} tr[N * 2], *pl;
void seg::init(long long l, long long r) {
  if (l == r) {
    a = in[l];
    return;
  }
  long long m = (l + r) / 2;
  lc = ++pl;
  lc->init(l, m);
  rc = ++pl;
  rc->init(m + 1, r);
  up();
}
void init(long long l, long long r) {
  pl = tr;
  tr->init(l, r);
}
void Solve() {
  long long nn = read(), q = read();
  long long n = 1 << nn;
  for (long long i = (0), _ = (n); i < _; i++) in[i] = read();
  init(0, n - 1);
  while (q--) {
    long long op = read();
    if (op == 1) {
      long long x = read() - 1, y = read();
      tr->change(0, n - 1, x, y);
    } else if (op == 2) {
      long long x = read();
      x = nn - x;
      tr->update(0, n - 1, 1 << x, 0);
    } else if (op == 3) {
      long long x = read();
      x = max(0ll, (nn - x) - 1);
      tr->update(0, n - 1, 0, 1 << x);
    } else {
      long long x = read() - 1, y = read() - 1;
      cout << tr->query(0, n - 1, x, y) << endl;
    }
  }
}
signed main() {
  long long T = 1;
  while (T--) Solve();
  return 0;
}
