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;
struct {
  long long val;
  int next;
} node[110000];
long long bit[63];
int a[63], b[63], head[63];
int n, cnt;
char str[63];
void init() {
  for (long long i = 0, j = 1; i < 63; ++i, j <<= 1) bit[(int)i] = j;
}
void clear() { cnt = 0; }
void insert(int x, long long val) {
  ++cnt;
  node[cnt].val = val;
  node[cnt].next = head[x];
  head[x] = cnt;
}
bool empty(int x) { return !head[x]; }
long long consume(int x) {
  long long ret = node[head[x]].val;
  head[x] = node[head[x]].next;
  return ret;
}
void work() {
  for (int i = 0; i < 63; ++i)
    if (a[i] > b[i] + 1) {
      printf("No\n");
      return;
    }
  printf("Yes\n");
  long long val = 0;
  while (1) {
    for (int i = 0; i < 63; ++i) {
      if (i == 63 - 1) {
        printf("\n");
        return;
      }
      if (!empty(i) && !(val & bit[i])) {
        long long next = consume(i);
        cout << next << " ";
        val ^= next;
        break;
      }
    }
  }
}
int main() {
  init();
  while (scanf("%d", &n) != EOF) {
    clear();
    memset(head, 0, sizeof(head));
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    long long val;
    for (int i = 0; i < n; ++i) {
      scanf("%s", str);
      val = 0;
      for (int i = 0, j = strlen(str); i < j; ++i)
        val = val * 10 + str[i] - '0';
      for (int i = 0; i < 63; ++i)
        if (val & bit[i]) {
          if (val - bit[i] > bit[i])
            b[i]++;
          else {
            a[i]++;
            insert(i, val);
            break;
          }
        }
    }
    work();
  }
}
