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;
template <class T>
void test(T a) {
  cout << a << endl;
}
template <class T, class T2>
void test(T a, T2 b) {
  cout << a << " " << b << endl;
}
template <class T, class T2, class T3>
void test(T a, T2 b, T3 c) {
  cout << a << " " << b << " " << c << endl;
}
const int N = 9e8 + 1;
const int M = 1e4 + 10;
const int MAXN = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const long long mod = (long long)1e9 + 7;
int T;
void testcase() { printf("Case #%d:\n", ++T); }
queue<long long> q[65];
long long a[MAXN];
long long ans[MAXN];
void work() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    long long x;
    scanf("%lld", &a[i]);
  }
  sort(a, a + n);
  for (int i = 1; i <= (n - 1); i++) {
    for (int j = 61; j >= 0; --j) {
      if ((1LL << j) & a[i]) {
        q[j].push(a[i]);
        break;
      }
    }
  }
  long long now = a[0];
  ans[0] = now;
  for (int i = 1; i <= (n - 1); i++) {
    bool flag = false;
    for (int j = 0; j <= 61; ++j) {
      if (((1LL << j) & now) == 0 && !q[j].empty()) {
        flag = true;
        ans[i] = q[j].front();
        now ^= q[j].front();
        q[j].pop();
        break;
      }
    }
    if (!flag) {
      printf("No\n");
      return;
    }
  }
  printf("Yes\n");
  for (int i = 0; i < (n); i++) {
    printf("%lld ", ans[i]);
  }
}
int main() { work(); }
