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>
long long q_pow(long long x, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = x * ret % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
long long __gcd(long long x, long long y) {
  if (!y) return x;
  return __gcd(y, x % y);
}
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
using namespace std;
const long long INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 50;
queue<long long> que[70];
long long ans[maxn];
signed main(int argc, char *argv[]) {
  if (argc == 2 && strcmp("-debug", argv[1]) == 0) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  int n;
  scanf("%d", &n);
  long long tmp;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &tmp);
    for (int j = 60; j >= 0; j--)
      if (tmp >> j & 1) {
        que[j].push(tmp);
        break;
      }
  }
  long long now = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 60; j++) {
      if ((now >> j & 1) == 0 && !que[j].empty()) {
        ans[i] = que[j].front();
        que[j].pop();
        now ^= ans[i];
        break;
      }
    }
  }
  int flag = 0;
  for (int i = 0; i <= 60; i++) flag = max(flag, (int)que[i].size());
  if (flag != 0) {
    cout << "No" << endl;
  } else {
    printf("Yes\n");
    for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  }
  return 0;
}
