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 long long maxn = 1e5 + 5;
const long long inf = 1e17 + 7;
const long long logg = 62;
long long hi(long long a) {
  if (a == 1) {
    return 0;
  }
  return hi(a >> 1) + 1;
}
vector<long long> b[logg];
long long nxt[maxn];
long long xo[maxn];
long long joda[maxn];
long long cnt = 1;
void pri() {
  long long be = nxt[0];
  while (be != -1) {
    cout << joda[be] << ' ';
    be = nxt[be];
  }
  cout << endl;
  long long h = 0;
  be = nxt[0];
  while (be != -1) {
    h ^= joda[be];
    be = nxt[be];
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long v;
    cin >> v;
    b[hi(v)].push_back(v);
  }
  nxt[0] = -1;
  xo[0] = 0;
  joda[0] = 0;
  for (long long i = logg - 1; i >= 0; i--) {
    long long be = 0;
    bool hal = 0;
    while (be != -1 && b[i].size()) {
      if ((((xo[be] >> i) ^ hal) & 1) == 0) {
        nxt[cnt] = nxt[be];
        nxt[be] = cnt;
        joda[cnt] = b[i].back();
        b[i].pop_back();
        be = cnt;
        cnt++;
        hal ^= 1;
      }
      be = nxt[be];
    }
    if (b[i].size()) {
      cout << "No";
      return 0;
    }
    be = 0;
    long long h = 0;
    while (be != -1) {
      h ^= joda[be];
      xo[be] = h;
      be = nxt[be];
    }
  }
  cout << "Yes\n";
  pri();
}
