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 maxn = 1e5 + 100;
int nxt[maxn * 60][2], tot;
int cnt[maxn * 60];
int flag[maxn * 60];
int n;
long long b[maxn];
long long a[maxn];
long long bas[100];
void insert(long long val) {
  int now = 0;
  for (int i = 61; i >= 0; i--) {
    long long bit = (val & bas[i]);
    if (bit) bit = 1;
    cnt[now]++;
    if (!nxt[now][bit]) nxt[now][bit] = ++tot;
    now = nxt[now][bit];
  }
  cnt[now]++;
}
long long query(long long Val, int dep, int now, bool same, long long ans) {
  if (dep == -1 && !same)
    return ans;
  else if (dep == -1 && same)
    return -1;
  long long bit = Val & bas[dep];
  if (bit) bit = 1;
  if (!same) {
    if (nxt[now][bit] && cnt[nxt[now][bit]]) {
      long long temp = query(Val, dep - 1, nxt[now][bit], same, ans);
      if (temp != -1) return temp;
    }
    if (nxt[now][!bit] && cnt[nxt[now][!bit]])
      return query(Val, dep - 1, nxt[now][!bit], same, ans + bas[dep]);
    else
      return -1;
  } else {
    if (bit) {
      if (nxt[now][!bit] && cnt[nxt[now][!bit]]) {
        long long temp =
            query(Val, dep - 1, nxt[now][!bit], same, ans + bas[dep]);
        if (temp != -1)
          return temp;
        else
          return -1;
      } else {
        return -1;
      }
    } else {
      if (nxt[now][bit] && cnt[nxt[now][bit]]) {
        long long temp = query(Val, dep - 1, nxt[now][bit], same, ans);
        if (temp != -1) return temp;
      }
      if (nxt[now][!bit] && cnt[nxt[now][!bit]])
        return query(Val, dep - 1, nxt[now][!bit], false, ans + bas[dep]);
      return -1;
    }
  }
}
void modify(long long Val) {
  int now = 0;
  for (int i = 61; i >= 0; i--) {
    cnt[now]--;
    long long bit = Val & bas[i];
    if (bit) bit = 1;
    now = nxt[now][bit];
  }
  cnt[now]--;
}
int main() {
  bas[0] = 1;
  for (int i = 1; i <= 63; i++) {
    bas[i] = bas[i - 1] << 1;
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", b + i);
    insert(b[i]);
  }
  for (int i = 1; i <= n; i++) {
    long long temp = query(a[i - 1], 61, 0, true, 0);
    if (temp == -1) {
      printf("No\n");
      return 0;
    } else {
      a[i] = temp;
      modify(a[i] ^ a[i - 1]);
    }
  }
  printf("Yes\n");
  for (int i = 1; i <= n; i++) {
    printf("%lld ", a[i] ^ a[i - 1]);
  }
  return 0;
}
