Vitya has learned that the answer for The Ultimate Question of Life, the Universe, and Everything is not the integer 54 42, but an increasing integer sequence a_1, …, a_n. In order to not reveal the secret earlier than needed, Vitya encrypted the answer and obtained the sequence b_1, …, b_n using the following rules:

  * b_1 = a_1;
  * b_i = a_i ⊕ a_{i - 1} for all i from 2 to n, where x ⊕ y is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of x and y. 



It is easy to see that the original sequence can be obtained using the rule a_i = b_1 ⊕ … ⊕ b_i.

However, some time later Vitya discovered that the integers b_i in the cypher got shuffled, and it can happen that when decrypted using the rule mentioned above, it can produce a sequence that is not increasing. In order to save his reputation in the scientific community, Vasya decided to find some permutation of integers b_i so that the sequence a_i = b_1 ⊕ … ⊕ b_i is strictly increasing. Help him find such a permutation or determine that it is impossible.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5).

The second line contains n integers b_1, …, b_n (1 ≤ b_i < 2^{60}).

Output

If there are no valid permutations, print a single line containing "No".

Otherwise in the first line print the word "Yes", and in the second line print integers b'_1, …, b'_n — a valid permutation of integers b_i. The unordered multisets \\{b_1, …, b_n\} and \\{b'_1, …, b'_n\} should be equal, i. e. for each integer x the number of occurrences of x in the first multiset should be equal to the number of occurrences of x in the second multiset. Apart from this, the sequence a_i = b'_1 ⊕ … ⊕ b'_i should be strictly increasing.

If there are multiple answers, print any of them.

Examples

Input

3
1 2 3


Output

No


Input

6
4 7 7 12 31 61


Output

Yes
4 12 7 31 7 61 

Note

In the first example no permutation is valid.

In the second example the given answer lead to the sequence a_1 = 4, a_2 = 8, a_3 = 15, a_4 = 16, a_5 = 23, a_6 = 42.
#include <bits/stdc++.h>
using namespace std;
int GetBit(long long x, int bit) { return (x >> bit) & 1; }
long long GetMask(int count) { return (1LL << count) - 1; }
const int BITS_COUNT = 60;
int n;
vector<long long> b;
bool Solve(const vector<long long>& b, const int bitId, vector<long long>& r) {
  if (bitId == BITS_COUNT) {
    assert(b.size() == 0);
    return true;
  }
  vector<long long> ones;
  vector<long long> nb;
  for (const auto bi : b) {
    const long long x = bi - (bi & GetMask(bitId));
    if (x == (1LL << bitId)) {
      ones.push_back(bi);
    } else {
      nb.push_back(bi);
    }
  }
  vector<long long> nr;
  if (!Solve(nb, bitId + 1, nr)) {
    return false;
  }
  if (ones.empty()) {
    r = nr;
    return true;
  }
  r.reserve(b.size());
  r.push_back(ones[0]);
  int oneId = 1;
  for (int i = 0; i < (int)nr.size(); ++i) {
    r.push_back(nr[i]);
    if (GetBit(nr[i], bitId) == 1) {
      if (oneId < (int)ones.size()) {
        r.push_back(ones[oneId++]);
      }
    }
  }
  if (oneId != (int)ones.size()) {
    return false;
  }
  return true;
}
void Solve() {
  vector<long long> a;
  if (!Solve(b, 0, a)) {
    cout << "No" << endl;
  } else {
    cout << "Yes" << endl;
    for (const auto x : a) {
      cout << x << " ";
    }
    cout << endl;
  }
}
bool Read() {
  if (!(cin >> n)) {
    return false;
  }
  b.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  while (Read()) {
    Solve();
  }
  return 0;
}
