#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1048579;
const long long SQRTN = 1003;
const long long LOGN = 22;
const double PI = acos(-1);
const long long INF = 1e16;
const long long MOD = 1000000007;
const long long FMOD = 998244353;
const double eps = 1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
long long add(long long a, long long b, long long c = MOD) {
  long long res = a + b;
  return (res >= c ? res - c : res);
}
long long mod_neg(long long a, long long b, long long c = MOD) {
  long long res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
long long mul(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return (res >= c ? res % c : res);
}
long long muln(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return ((res % c) + c) % c;
}
long long mulmod(long long a, long long b, long long m = MOD) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m = MOD) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mul(x, p, m);
    p = mul(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T& x, T& y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n = MOD) {
  T x, y, z = 0;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : mod_neg(x, z, n));
}
const long long FACSZ = 1;
long long fact[FACSZ], ifact[FACSZ];
void precom(long long c = MOD) {
  fact[0] = 1;
  for (long long i = 1; i < FACSZ; i++) fact[i] = mul(fact[i - 1], i, c);
  ifact[FACSZ - 1] = mod_inverse(fact[FACSZ - 1], c);
  for (long long i = FACSZ - 1 - 1; i >= 0; i--) {
    ifact[i] = mul(i + 1, ifact[i + 1], c);
  }
}
long long ncr(long long n, long long r, long long c = MOD) {
  if (r > n) return 0;
  return mul(mul(ifact[r], ifact[n - r], c), fact[n], c);
}
void solvethetestcase();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long testcase = 1; testcase < t + 1; testcase++) {
    solvethetestcase();
  }
}
long long n, k;
long long p2;
struct node {
  long long mn, mx, val, ln, lft, rgt;
  node() {
    mn = INF;
    mx = -INF;
    val = INF;
    lft = -1;
    rgt = -1;
    ln = 1;
  }
};
node seg[MAXN];
long long ans[MAXN];
void calc(long long i) {
  long long l = seg[i].ln / 2;
  seg[i].mn = min(seg[seg[i].lft].mn, seg[seg[i].rgt].mn + l);
  seg[i].mx = max(seg[seg[i].lft].mx, seg[seg[i].rgt].mx + l);
  seg[i].val = min({seg[2 * i].val, seg[2 * i + 1].val,
                    l + seg[seg[i].rgt].mn - seg[seg[i].lft].mx});
}
void solvethetestcase() {
  cin >> n >> k;
  p2 = (1 << k);
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    seg[x + p2].mn = 0;
    seg[x + p2].mx = 0;
  }
  for (long long i = p2 - 1; i >= 1; i--) {
    seg[i].lft = 2 * i;
    seg[i].rgt = 2 * i + 1;
    seg[i].ln = seg[2 * i].ln * 2;
    calc(i);
  }
  ans[0] = seg[1].val;
  long long cur = 0;
  for (long long i = 1; i < (1 << k); i++) {
    long long hp = 0;
    for (long long j = 0; j < k; j++) {
      if (i % (1 << j) == 0)
        hp = j;
      else
        break;
    }
    cur ^= (1 << (k - 1 - hp));
    for (long long j = (1 << hp); j < (1 << (hp + 1)); j++) {
      swap(seg[j].lft, seg[j].rgt);
      calc(j);
    }
    for (long long j = (1 << hp) - 1; j >= 1; j--) calc(j);
    ans[cur] = seg[1].val;
  }
  for (long long i = 0; i < (1 << k); i++) cout << ans[i] << " ";
  cout << "\n";
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

You are given an array a consisting of n distinct elements and an integer k. Each element in the array is a non-negative integer not exceeding 2^k-1.

Let's define the XOR distance for a number x as the value of 

$$$f(x) = min_{i = 1}^{n} min_{j = i + 1}^{n} |(a_i ⊕ x) - (a_j ⊕ x)|,$$$

where ⊕ denotes [the bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

For every integer x from 0 to 2^k-1, you have to calculate f(x).

Input

The first line contains two integers n and k (1 ≤ k ≤ 19; 2 ≤ n ≤ 2^k).

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^k-1). All these integers are distinct.

Output

Print 2^k integers. The i-th of them should be equal to f(i-1).

Examples

Input


3 3
6 0 3


Output


3 1 1 2 2 1 1 3 


Input


3 4
13 4 2


Output


2 2 6 6 3 1 2 2 2 2 1 3 6 6 2 2 

Note

Consider the first example:

  * for x = 0, if we apply bitwise XOR to the elements of the array with x, we get the array [6, 0, 3], and the minimum absolute difference of two elements is 3; 
  * for x = 1, if we apply bitwise XOR to the elements of the array with x, we get the array [7, 1, 2], and the minimum absolute difference of two elements is 1; 
  * for x = 2, if we apply bitwise XOR to the elements of the array with x, we get the array [4, 2, 1], and the minimum absolute difference of two elements is 1; 
  * for x = 3, if we apply bitwise XOR to the elements of the array with x, we get the array [5, 3, 0], and the minimum absolute difference of two elements is 2; 
  * for x = 4, if we apply bitwise XOR to the elements of the array with x, we get the array [2, 4, 7], and the minimum absolute difference of two elements is 2; 
  * for x = 5, if we apply bitwise XOR to the elements of the array with x, we get the array [3, 5, 6], and the minimum absolute difference of two elements is 1; 
  * for x = 6, if we apply bitwise XOR to the elements of the array with x, we get the array [0, 6, 5], and the minimum absolute difference of two elements is 1; 
  * for x = 7, if we apply bitwise XOR to the elements of the array with x, we get the array [1, 7, 4], and the minimum absolute difference of two elements is 3. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1048579;
const long long SQRTN = 1003;
const long long LOGN = 22;
const double PI = acos(-1);
const long long INF = 1e16;
const long long MOD = 1000000007;
const long long FMOD = 998244353;
const double eps = 1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
long long add(long long a, long long b, long long c = MOD) {
  long long res = a + b;
  return (res >= c ? res - c : res);
}
long long mod_neg(long long a, long long b, long long c = MOD) {
  long long res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
long long mul(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return (res >= c ? res % c : res);
}
long long muln(long long a, long long b, long long c = MOD) {
  long long res = (long long)a * b;
  return ((res % c) + c) % c;
}
long long mulmod(long long a, long long b, long long m = MOD) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m = MOD) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mul(x, p, m);
    p = mul(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T& x, T& y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n = MOD) {
  T x, y, z = 0;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : mod_neg(x, z, n));
}
const long long FACSZ = 1;
long long fact[FACSZ], ifact[FACSZ];
void precom(long long c = MOD) {
  fact[0] = 1;
  for (long long i = 1; i < FACSZ; i++) fact[i] = mul(fact[i - 1], i, c);
  ifact[FACSZ - 1] = mod_inverse(fact[FACSZ - 1], c);
  for (long long i = FACSZ - 1 - 1; i >= 0; i--) {
    ifact[i] = mul(i + 1, ifact[i + 1], c);
  }
}
long long ncr(long long n, long long r, long long c = MOD) {
  if (r > n) return 0;
  return mul(mul(ifact[r], ifact[n - r], c), fact[n], c);
}
void solvethetestcase();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long testcase = 1; testcase < t + 1; testcase++) {
    solvethetestcase();
  }
}
long long n, k;
long long p2;
struct node {
  long long mn, mx, val, ln, lft, rgt;
  node() {
    mn = INF;
    mx = -INF;
    val = INF;
    lft = -1;
    rgt = -1;
    ln = 1;
  }
};
node seg[MAXN];
long long ans[MAXN];
void calc(long long i) {
  long long l = seg[i].ln / 2;
  seg[i].mn = min(seg[seg[i].lft].mn, seg[seg[i].rgt].mn + l);
  seg[i].mx = max(seg[seg[i].lft].mx, seg[seg[i].rgt].mx + l);
  seg[i].val = min({seg[2 * i].val, seg[2 * i + 1].val,
                    l + seg[seg[i].rgt].mn - seg[seg[i].lft].mx});
}
void solvethetestcase() {
  cin >> n >> k;
  p2 = (1 << k);
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    seg[x + p2].mn = 0;
    seg[x + p2].mx = 0;
  }
  for (long long i = p2 - 1; i >= 1; i--) {
    seg[i].lft = 2 * i;
    seg[i].rgt = 2 * i + 1;
    seg[i].ln = seg[2 * i].ln * 2;
    calc(i);
  }
  ans[0] = seg[1].val;
  long long cur = 0;
  for (long long i = 1; i < (1 << k); i++) {
    long long hp = 0;
    for (long long j = 0; j < k; j++) {
      if (i % (1 << j) == 0)
        hp = j;
      else
        break;
    }
    cur ^= (1 << (k - 1 - hp));
    for (long long j = (1 << hp); j < (1 << (hp + 1)); j++) {
      swap(seg[j].lft, seg[j].rgt);
      calc(j);
    }
    for (long long j = (1 << hp) - 1; j >= 1; j--) calc(j);
    ans[cur] = seg[1].val;
  }
  for (long long i = 0; i < (1 << k); i++) cout << ans[i] << " ";
  cout << "\n";
}

```