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;
#pragma comment(linker, "/STACK:102400000,102400000")
const int maxn = 1e5 + 5;
const long long inf = (1 << 62);
int n;
vector<long long> ve;
vector<long long> bite;
vector<long long> G[maxn];
int pos[maxn];
int maxlen;
void get_bite(long long sum) {
  bite.clear();
  while (sum) {
    bite.push_back(sum % 2);
    sum >>= 1;
  }
}
bool solve() {
  long long sum = 0;
  vector<long long> ans;
  for (int i = 1; i <= n; i++) {
    get_bite(sum);
    bool flag = false;
    for (int j = 1; j <= maxlen; j++) {
      if (j == bite.size())
        continue;
      else if (j < bite.size() && (bite[j - 1] == 0)) {
        if (G[j].size() > 0 && (G[j].size() > pos[j])) {
          flag = true;
          ans.push_back(G[j][pos[j]]);
          sum ^= G[j][pos[j]];
          pos[j]++;
          break;
        }
      } else if (j > bite.size()) {
        if (G[j].size() > 0 && (G[j].size() > pos[j])) {
          flag = true;
          ans.push_back(G[j][pos[j]]);
          sum ^= G[j][pos[j]];
          pos[j]++;
          break;
        }
      }
    }
    if (!flag) return false;
  }
  puts("Yes");
  for (int i = 0; i < ans.size(); i++) {
    printf("%lld ", ans[i]);
  }
  cout << endl;
  return true;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long a;
    scanf("%lld", &a);
    ve.push_back(a);
  }
  for (int i = 0; i < ve.size(); i++) {
    long long tmp = ve[i];
    int cnt = 0;
    while (tmp) {
      cnt++;
      tmp >>= 1;
    }
    G[cnt].push_back(ve[i]);
    maxlen = max(maxlen, cnt);
  }
  for (int i = 1; i <= maxlen; i++) {
    sort(G[i].begin(), G[i].end());
  }
  if (solve() == false) puts("No");
  return 0;
}
