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;
const int maxn = (int)1e5 + 100;
const int mod = (int)1e9 + 7;
const int P = (int)1e6 + 7;
const double pi = acos(-1.0);
int n;
long long a[maxn];
vector<int> g[70];
bool cmp(int i, int j) { return a[i] < a[j]; }
void solve() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    scanf("%lld", &a[i]);
    for (int b = (59); b >= (0); --b) {
      if ((a[i] >> b) & 1) {
        g[b].push_back(i);
        break;
      }
    }
  }
  vector<int> ans;
  for (int i = (59); i >= (0); --i) {
    vector<int> nxt;
    int ptr = 0;
    sort((g[i]).begin(), (g[i]).end(), cmp);
    long long cur = 0;
    for (auto x : g[i]) {
      if (cur < (cur ^ a[x]))
        nxt.push_back(x), cur ^= a[x];
      else {
        while (ptr < ((int)(ans).size()) && cur >= (cur ^ a[x])) {
          cur ^= a[ans[ptr]];
          nxt.push_back(ans[ptr++]);
        }
        if (cur >= (cur ^ a[x])) {
          puts("No");
          return;
        }
        cur ^= a[x];
        nxt.push_back(x);
      }
    }
    while (ptr < ((int)(ans).size())) nxt.push_back(ans[ptr++]);
    swap(ans, nxt);
  }
  puts("Yes");
  for (auto x : ans) printf("%lld ", a[x]);
}
int main() {
  srand(time(0));
  int t = 1;
  while (t--) solve();
}
