#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
long long power(long long x, long long y, long long M) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % M;
    y >>= 1;
    x = (x * x) % M;
  }
  return res % M;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y >>= 1;
    x = (x * x);
  }
  return res;
}
const long long mxn = 1e6 + 6, mod = 1e9 + 7, inf = 1e18;
long long t, casee = 1;
void test_case() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> possiblesubsetxor(9e3, vector<long long>());
  for (long long i = 0; i < 9e3; i++) possiblesubsetxor[i].push_back(0);
  vector<long long> pos(9e3, 0);
  vector<long long> help(9e3, 9e3);
  pos[0] = 1;
  help[0] = 0;
  for (long long i = 0; i < n; i++) {
    for (long long currxor : possiblesubsetxor[a[i]]) {
      long long newxor = currxor ^ a[i];
      pos[newxor] = 1;
      if (a[i] < help[newxor]) {
        for (long long k = a[i] + 1; k < help[newxor]; k++) {
          possiblesubsetxor[k].push_back(newxor);
        }
        help[newxor] = a[i];
      }
    }
    possiblesubsetxor[a[i]].clear();
  }
  long long cnt = 0;
  for (long long i = 0; i < 9e3; i++) cnt += pos[i];
  cout << cnt << "\n";
  for (long long i = 0; i < 9e3; i++) {
    if (pos[i]) cout << i << " ";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    test_case();
    casee++;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

This is a harder version of the problem with bigger constraints.

Korney Korneevich dag up an array a of length n. Korney Korneevich has recently read about the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), so he wished to experiment with it. For this purpose, he decided to find all integers x ≥ 0 such that there exists an increasing subsequence of the array a, in which the bitwise XOR of numbers is equal to x.

It didn't take a long time for Korney Korneevich to find all such x, and he wants to check his result. That's why he asked you to solve this problem!

A sequence s is a subsequence of a sequence b if s can be obtained from b by deletion of several (possibly, zero or all) elements.

A sequence s_1, s_2, … , s_m is called increasing if s_1 < s_2 < … < s_m.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^6).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 5000) — the elements of the array a.

Output

In the first line print a single integer k — the number of found x values.

In the second line print k integers in increasing order x_1, x_2, … x_k (0 ≤ x_1 < … < x_k) — found x values.

Examples

Input


4
4 2 2 4


Output


4
0 2 4 6 


Input


8
1 0 1 7 12 5 3 2


Output


12
0 1 2 3 4 5 6 7 10 11 12 13 

Note

In the first test case:

  * To get value x = 0 it is possible to choose and empty subsequence 
  * To get value x = 2 it is possible to choose a subsequence [2] 
  * To get value x = 4 it is possible to choose a subsequence [4] 
  * To get value x = 6 it is possible to choose a subsequence [2, 4] 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
long long power(long long x, long long y, long long M) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % M;
    y >>= 1;
    x = (x * x) % M;
  }
  return res % M;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y >>= 1;
    x = (x * x);
  }
  return res;
}
const long long mxn = 1e6 + 6, mod = 1e9 + 7, inf = 1e18;
long long t, casee = 1;
void test_case() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> possiblesubsetxor(9e3, vector<long long>());
  for (long long i = 0; i < 9e3; i++) possiblesubsetxor[i].push_back(0);
  vector<long long> pos(9e3, 0);
  vector<long long> help(9e3, 9e3);
  pos[0] = 1;
  help[0] = 0;
  for (long long i = 0; i < n; i++) {
    for (long long currxor : possiblesubsetxor[a[i]]) {
      long long newxor = currxor ^ a[i];
      pos[newxor] = 1;
      if (a[i] < help[newxor]) {
        for (long long k = a[i] + 1; k < help[newxor]; k++) {
          possiblesubsetxor[k].push_back(newxor);
        }
        help[newxor] = a[i];
      }
    }
    possiblesubsetxor[a[i]].clear();
  }
  long long cnt = 0;
  for (long long i = 0; i < 9e3; i++) cnt += pos[i];
  cout << cnt << "\n";
  for (long long i = 0; i < 9e3; i++) {
    if (pos[i]) cout << i << " ";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    test_case();
    casee++;
  }
  return 0;
}

```