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;
int ch[15100000][2], sum[15100000], cnt, root[65], n;
long long b[301000], bin[65], c[301000];
void insert(int &, long long);
long long query(int, long long);
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= 62; i++) bin[i] = 1ll << i;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &b[i]);
    for (int j = 60; j >= 0; j--) {
      if (b[i] & bin[j]) {
        insert(root[j], b[i]);
        break;
      }
    }
  }
  long long zz = 0;
  int t = 1;
  for (; t <= n; t++) {
    for (int i = 0; i <= 60; i++) {
      if ((!(zz & bin[i])) && sum[root[i]] != 0) {
        c[t] = query(root[i], zz);
        zz ^= c[t];
        break;
      }
    }
    if (!c[t]) break;
  }
  if (t <= n)
    printf("No");
  else {
    printf("Yes\n");
    for (int i = 1; i <= n; i++) printf("%lld ", c[i]);
  }
  return 0;
}
void insert(int &rt, long long y) {
  if (!rt) rt = ++cnt;
  sum[rt]++;
  int now = rt;
  for (int i = 60; i >= 0; i--) {
    if (y & bin[i]) {
      if (!ch[now][1]) ch[now][1] = ++cnt;
      now = ch[now][1];
    } else {
      if (!ch[now][0]) ch[now][0] = ++cnt;
      now = ch[now][0];
    }
    sum[now]++;
  }
}
long long query(int x, long long y) {
  sum[x]--;
  int now = x;
  long long ans = 0;
  for (int i = 60; i >= 0; i--) {
    int d = ((y & bin[i]) != 0);
    if (sum[ch[now][d]]) {
      now = ch[now][d];
      if (d) ans += bin[i];
    } else {
      now = ch[now][d ^ 1];
      if (!d) ans += bin[i];
    }
    sum[now]--;
  }
  return ans;
}
