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;
inline long long read();
const int N = 1e5 + 2;
int n, tp, t[N];
long long x[N], sum;
queue<int> c[62];
void Work() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    x[i] = read();
    sum ^= x[i];
    for (int j = 60; j >= 0; --j)
      if ((x[i] >> j) & 1) {
        c[j].push(i);
        break;
      }
  }
  while (sum) {
    bool ok = 0;
    for (int i = 0; i <= 60; ++i)
      if ((sum >> i) & 1 && !c[i].empty()) {
        t[++tp] = c[i].front();
        c[i].pop();
        sum ^= x[t[tp]];
        ok = 1;
        break;
      }
    if (!ok) {
      printf("No");
      return;
    }
  }
  if (tp != n) {
    printf("No");
    return;
  }
  printf("Yes\n");
  for (int i = n; i >= 1; --i) printf("%lld ", x[t[i]]);
}
int main() {
  Work();
  return 0;
}
inline long long read() {
  char c;
  bool type = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') type = 0;
  long long ans = c ^ 48;
  while ((c = getchar()) >= '0' && c <= '9')
    ans = (ans << 3) + (ans << 1) + (c ^ 48);
  return type ? ans : -ans;
}
