#include <bits/stdc++.h>
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + (c - '0'), c = getchar();
  return f * x;
}
long long min(long long a, long long b) { return a < b ? a : b; }
long long max(long long a, long long b) { return a > b ? a : b; }
template <typename T>
bool umin(T& a, T t) {
  if (a > t) return a = t, 1;
  return 0;
}
template <typename T>
bool umax(T& a, T t) {
  if (a < t) return a = t, 1;
  return 0;
}
const int MAXN = 5011, B = 85;
typedef std::bitset<(1 << 13)> bits;
bits a[MAXN], nxt[MAXN], res;
std::vector<short> seq[MAXN];
int main() {
  int n = read();
  for (int i = 0; i <= 5000; ++i) nxt[i][0] = 1, seq[i].emplace_back(0);
  a[0][0] = 1;
  while (n--) {
    int x = read();
    for (auto y : seq[x]) {
      int k = y ^ x;
      a[x][k] = 1;
      for (int i = x + 1; i <= 5000; ++i)
        if (!nxt[i][k])
          nxt[i][k] = 1, seq[i].emplace_back(k);
        else
          break;
    }
    seq[x].clear();
  }
  for (int i = 0; i <= 5000; ++i) res |= a[i];
  printf("%d\n", int(res.count()));
  for (int i = 0; i < (1 << 13); ++i)
    if (res[i]) printf("%d ", i);
  puts("");
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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>
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + (c - '0'), c = getchar();
  return f * x;
}
long long min(long long a, long long b) { return a < b ? a : b; }
long long max(long long a, long long b) { return a > b ? a : b; }
template <typename T>
bool umin(T& a, T t) {
  if (a > t) return a = t, 1;
  return 0;
}
template <typename T>
bool umax(T& a, T t) {
  if (a < t) return a = t, 1;
  return 0;
}
const int MAXN = 5011, B = 85;
typedef std::bitset<(1 << 13)> bits;
bits a[MAXN], nxt[MAXN], res;
std::vector<short> seq[MAXN];
int main() {
  int n = read();
  for (int i = 0; i <= 5000; ++i) nxt[i][0] = 1, seq[i].emplace_back(0);
  a[0][0] = 1;
  while (n--) {
    int x = read();
    for (auto y : seq[x]) {
      int k = y ^ x;
      a[x][k] = 1;
      for (int i = x + 1; i <= 5000; ++i)
        if (!nxt[i][k])
          nxt[i][k] = 1, seq[i].emplace_back(k);
        else
          break;
    }
    seq[x].clear();
  }
  for (int i = 0; i <= 5000; ++i) res |= a[i];
  printf("%d\n", int(res.count()));
  for (int i = 0; i < (1 << 13); ++i)
    if (res[i]) printf("%d ", i);
  puts("");
  return 0;
}

```