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 = 1e5 + 50;
long long b[N], a[N], vec[62][N], cnt[62], c[62];
long long lowbit(long long x) { return x & (-x); }
map<long long, int> mp;
int main() {
  int n;
  for (int i = 0; i <= 60; i++) mp[(1LL << i)] = i;
  memset(cnt, 0, sizeof(cnt));
  memset(c, 0, sizeof(c));
  scanf("%d", &n);
  a[n - 1] = 0;
  long long val, len, x;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &b[i]);
    a[n - 1] ^= b[i];
    val = b[i], len = 0;
    while (val) {
      x = lowbit(val);
      len = max(len, (long long)mp[x]);
      val -= x;
    }
    vec[len][cnt[len]++] = b[i];
  }
  for (int i = n - 1; i > 0; i--) {
    if (a[i] == 0) {
      printf("No\n");
      return 0;
    }
    val = a[i];
    bool flag = false;
    while (val) {
      x = lowbit(val);
      len = mp[x];
      if (c[len] >= cnt[len]) {
        val -= x;
        continue;
      }
      a[i - 1] = a[i] ^ vec[len][c[len]++];
      flag = true;
      break;
    }
    if (!flag) {
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  printf("%lld ", a[0]);
  for (int i = 1; i < n; i++) printf("%lld ", a[i] ^ a[i - 1]);
  puts("");
}
