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 N = 2e5 + 5;
const long long mod = 99998244353;
long long n, m, t, k, a[N], b[N], tong, ans, d[N], u, v, h[N], st[N], fe[N];
priority_queue<long long, vector<long long>, greater<long long> > pq;
struct dang {
  long long s, f, l, b;
} e[N];
vector<long long> adj[N], kq;
vector<pair<long long, long long> > w;
void add(long long id) {
  for (; id <= n; id += id & -id) st[id] += v;
}
long long get(long long id) {
  long long total = 0;
  for (; id; id -= id & -id) total += st[id];
  return total;
}
void addfe(long long id) {
  for (; id <= n; id += id & -id) fe[id] += v;
}
long long getfe(long long id) {
  long long total = 0;
  for (; id; id -= id & -id) total += fe[id];
  return total;
}
void sol() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 62; j >= 0; j--) {
      if ((a[i] >> j) & 1) {
        adj[j].push_back(a[i]);
        break;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    bool ok = false;
    for (int j = 0; j < 63; j++) {
      if (!((tong >> j) & 1) && adj[j].size()) {
        tong ^= adj[j].back();
        kq.push_back(adj[j].back());
        adj[j].pop_back();
        ok = true;
        break;
      }
    }
    if (!ok) {
      cout << "No";
      return;
    }
  }
  cout << "Yes" << '\n';
  for (long long u : kq) cout << u << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("haircut"
            ".in",
            "r")) {
    freopen(
        "haircut"
        ".in",
        "r", stdin);
    freopen(
        "haircut"
        ".out",
        "w", stdout);
  }
  int ntest = 1;
  while (ntest-- > 0) sol();
}
