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 inf = (int)1e9 + 7;
struct fraction {
  int p;
  int q;
  friend bool operator<(fraction a, fraction b);
};
bool operator<(fraction a, fraction b) {
  return (long long)a.p * b.q < (long long)a.q * b.p;
}
bool mutant_landfill(vector<pair<int, pair<int, int>>> &bohr,
                     vector<long long> &b, vector<bool> &used, int bit,
                     int &cnt, int v, long long &bitmask) {
  if (cnt >= b.size()) return true;
  if (bohr[v].second.first == -1 && !((bitmask >> bit) & 1)) {
    for (; cnt < b.size(); cnt++) {
      if (!used[cnt] && ((b[cnt] >> bit) & 1)) {
        used[cnt] = true;
        bohr[v].second.first = bohr.size();
        bohr.emplace_back(cnt, make_pair(-1, -1));
        bitmask ^= b[cnt];
        break;
      }
    }
  } else if (bohr[v].second.first != -1)
    mutant_landfill(bohr, b, used, bit, cnt, bohr[v].second.first, bitmask);
  bitmask ^= b[bohr[v].first];
  if (bohr[v].second.second == -1 && !((bitmask >> bit) & 1)) {
    for (; cnt < b.size(); cnt++) {
      if (!used[cnt] && ((b[cnt] >> bit) & 1)) {
        used[cnt] = true;
        bohr[v].second.second = bohr.size();
        bohr.emplace_back(cnt, make_pair(-1, -1));
        bitmask ^= b[cnt];
        break;
      }
    }
  } else if (bohr[v].second.second != -1)
    mutant_landfill(bohr, b, used, bit, cnt, bohr[v].second.second, bitmask);
  if (v == 0) {
    for (; cnt < b.size(); cnt++) {
      if (!used[cnt] && ((b[cnt] >> bit) & 1)) {
        return false;
      }
    }
    return true;
  }
}
void traverse(vector<pair<int, pair<int, int>>> &bohr, vector<long long> &b,
              int v) {
  if (bohr[v].second.first != -1) traverse(bohr, b, bohr[v].second.first);
  cout << b[bohr[v].first] << ' ';
  if (bohr[v].second.second != -1) traverse(bohr, b, bohr[v].second.second);
}
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int n;
  cin >> n;
  vector<long long> b(n);
  for (int i = 0; i < n; i++) cin >> b[i];
  vector<int> cnt_bits(60);
  for (int i = 0; i < n; i++) {
    for (int bit = 0; bit < 60; bit++) {
      cnt_bits[bit] += (b[i] >> bit) & 1;
    }
  }
  vector<bool> used(n);
  vector<pair<int, pair<int, int>>> bohr;
  for (int start = 59; start >= 0; start--) {
    if (cnt_bits[start] > 0) {
      if (cnt_bits[start] > 1) {
        cout << "No" << endl;
        return;
      } else {
        for (int i = 0; i < n; i++) {
          if ((b[i] >> start) & 1) {
            bohr.emplace_back(i, make_pair(-1, -1));
            used[i] = true;
          }
        }
        int cnt = 0;
        long long bitmask = 0;
        for (int bit = start - 1; bit >= 0; bit--) {
          if (!mutant_landfill(bohr, b, used, bit, cnt, 0, bitmask)) {
            cout << "No" << endl;
            return;
          }
          cnt = 0;
          bitmask = 0;
        }
        cout << "Yes" << endl;
        traverse(bohr, b, 0);
        cout << endl;
        return;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
