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 data {
  int val;
  struct data *next[2];
};
long long p2[63];
void pre() {
  p2[0] = 1LL;
  for (int i = 1; i < 60; i++) {
    p2[i] = p2[i - 1] * 2;
  }
  return;
}
struct data *new_mem() {
  struct data *baru = (struct data *)malloc(sizeof(struct data));
  baru->next[0] = NULL;
  baru->next[1] = NULL;
  baru->val = 0;
  return baru;
}
void add_trie(long long x, struct data **root) {
  struct data *cur = *root;
  for (int i = 59; i >= 0; i--) {
    long long now = x / p2[i];
    x %= p2[i];
    if (now == 0) {
      if (cur->next[0] == NULL) {
        cur->next[0] = new_mem();
      }
      cur = cur->next[0];
    } else {
      if (cur->next[1] == NULL) {
        cur->next[1] = new_mem();
      }
      cur = cur->next[1];
    }
    cur->val++;
  }
  return;
}
void delete_trie(long long x, struct data **root) {
  struct data *cur = *root;
  for (int i = 59; i >= 0; i--) {
    long long now = x / p2[i];
    x %= p2[i];
    if (now == 0) {
      cur = cur->next[0];
    } else {
      cur = cur->next[1];
    }
    cur->val--;
  }
  return;
}
long long dfs(long long x, struct data **root, int lvl) {
  if (lvl == -1) return 0LL;
  long long ret = 0LL;
  long long _on = x / p2[lvl];
  struct data *now = *root;
  if (_on == 1) {
    if (now->next[0] == NULL || (now->next[0])->val == 0) return -1;
    now = now->next[0];
    ret = dfs(x % p2[lvl], &now, lvl - 1);
    return ret;
  } else {
    if (now->next[0] != NULL && (now->next[0])->val != 0) {
      ret = dfs(x % p2[lvl], &(now->next[0]), lvl - 1);
      if (ret != -1) return ret;
    }
    if (now->next[1] != NULL && (now->next[1])->val != 0) {
      ret = dfs(0, &(now->next[1]), lvl - 1);
      if (ret != -1) return p2[lvl] + ret;
    }
    return ret;
  }
}
vector<long long> ans;
int n;
void solve(struct data **root) {
  long long now = 0LL;
  for (int i = 1; i <= n; i++) {
    long long cur = dfs(now, root, 59);
    if (cur == -1) return;
    ans.push_back(cur);
    delete_trie(cur, root);
    now = now ^ cur;
  }
  return;
}
long long x;
int main() {
  pre();
  struct data *root = new_mem();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &x);
    add_trie(x, &root);
  }
  solve(&root);
  if (ans.size() != n)
    printf("No\n");
  else {
    printf("Yes\n");
    for (int i = 0; i < n; i++) {
      if (i > 0) printf(" ");
      printf("%lld", ans[i]);
    }
  }
  return 0;
}
