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;
void qmax(long long &x, long long y) {
  if (x < y) x = y;
}
void qmin(long long &x, long long y) {
  if (x > y) x = y;
}
inline long long read() {
  char s;
  long long k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) {
    k = k * 10 + (s ^ '0');
    s = getchar();
  }
  return k * base;
}
inline void write(long long x) {
  static char cnt, num[15];
  cnt = 0;
  if (!x) {
    printf("0");
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
long long n, l[70], x, f[70], ans[2001010];
vector<long long> a[70];
long long flag;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) {
    x = read();
    for (long long i = 59; i >= 0; i--) {
      if (x & (1ll << i)) {
        a[i].push_back(x);
        l[i]++;
        break;
      }
    }
  }
  long long lst = 0;
  for (long long i = 0; i < 60; i++)
    if (l[i] != 0) sort(a[i].begin(), a[i].end());
  for (long long j = 1; j <= n; j++) {
    flag = 0;
    for (long long i = 0; i < 60; i++) {
      if (f[i] == l[i]) continue;
      if (((lst >> (long long)i) & 1ll) != 0) continue;
      f[i]++;
      ans[j] = a[i][f[i] - 1];
      flag = 1;
      lst ^= ans[j];
      break;
    }
    if (!flag) break;
  }
  if (!flag) {
    printf("No");
    return 0;
  }
  printf("Yes\n");
  for (long long i = 1; i <= n; i++) {
    printf("%lld ", ans[i]);
  }
  return 0;
}
