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 N = 2e6 + 229;
const long long M = 1e9 + 7;
const long long M2 = 1e9 + 9;
const long long mod = 998244353;
const int rx[8] = {1, 1, -1, -1, -2, -2, 2, 2};
const int ry[8] = {-2, 2, -2, 2, 1, -1, 1, -1};
const long double eps = 1e-7;
const double pi = acos(-1.0);
long long ot[N], a[N];
int sz = 0, k[4 * N], nxt[N * 4][2];
void add(long long x) {
  int v = 0, to = 0;
  for (int i = 59; i >= 0; --i) {
    to = 0;
    if ((1ll << i) & x) {
      to = 1;
    }
    if (!nxt[v][to]) {
      ++sz;
      nxt[v][to] = sz;
    }
    v = nxt[v][to];
    ++k[v];
  }
}
void del(long long x) {
  int v = 0, to = 0;
  for (int i = 59; i >= 0; --i) {
    to = 0;
    if ((1ll << i) & x) {
      to = 1;
    }
    v = nxt[v][to];
    --k[v];
  }
}
int find_bit(long long x) {
  int v = 0, to = 0;
  int ot = 1e9;
  for (int i = 59; i >= 0; --i) {
    to = 0;
    if (!((1ll << i) & x) && nxt[v][1] && k[nxt[v][1]]) {
      ot = min(ot, i);
    }
    if (!nxt[v][to]) {
      break;
    }
    v = nxt[v][to];
  }
  if (ot == 1e9) {
    ot = -1;
  }
  return ot;
}
long long solve(long long x) {
  int v = 0, to = 0, j = find_bit(x);
  long long ot = 0;
  if (j == -1) {
    return -1;
  }
  for (int i = 59; i >= 0; --i) {
    to = 0;
    if (i > j) {
      v = nxt[v][to];
    } else if (i == j) {
      ot += (1ll << i);
      v = nxt[v][1];
    } else {
      if ((1ll << i) & x) {
        to = 1;
      }
      if (!nxt[v][to] || !k[nxt[v][to]]) {
        to = !to;
      }
      if (to) {
        ot += (1ll << i);
      }
      v = nxt[v][to];
    }
  }
  return ot;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  int n, ind = 0;
  long long now = 0;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] < a[ind]) {
      ind = i;
    }
    add(a[i]);
  }
  ot[0] = a[ind];
  now = a[ind];
  del(a[ind]);
  for (int i = 1; i < n; ++i) {
    ot[i] = solve(now);
    del(ot[i]);
    if (ot[i] == -1) {
      cout << "No";
      return 0;
    }
    now ^= ot[i];
  }
  now = 0;
  for (int i = 0; i < n; ++i) {
    if (now >= (now ^ ot[i])) {
      cout << "No";
      return 0;
    }
    now ^= ot[i];
  }
  cout << "Yes" << '\n';
  for (int i = 0; i < n; ++i) {
    cout << ot[i] << ' ';
  }
  return 0;
}
