Problem: In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.

When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.

The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.

For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.

<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines. 

You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.

The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.

The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.

It is guaranteed that the sum of n across all test cases does not exceed 10^5.

Output

For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).

If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.

If there are multiple solutions, output any.

Example

Input


7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1


Output


YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1

Note

The first test case is described in the statement.

In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.

In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.

In the fourth test case, it can be proved that no solution exists.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using vpdd = vector<pdd>;
template <typename T>
void ckmin(T& a, const T& b) {
  a = min(a, b);
}
template <typename T>
void ckmax(T& a, const T& b) {
  a = max(a, b);
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
namespace __input {
template <class T1, class T2>
void re(pair<T1, T2>& p);
template <class T>
void re(vector<T>& a);
template <class T, size_t SZ>
void re(array<T, SZ>& a);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(double& x) {
  string t;
  re(t);
  x = stod(t);
}
template <class Arg, class... Args>
void re(Arg& first, Args&... rest) {
  re(first);
  re(rest...);
}
template <class T1, class T2>
void re(pair<T1, T2>& p) {
  re(p.first, p.second);
}
template <class T>
void re(vector<T>& a) {
  for (int i = 0; i < (int((a).size())); i++) re(a[i]);
}
template <class T, size_t SZ>
void re(array<T, SZ>& a) {
  for (int i = 0; i < (SZ); i++) re(a[i]);
}
}  // namespace __input
using namespace __input;
namespace __output {
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T, size_t SZ>
void pr(const array<T, SZ>& x);
template <class T>
void pr(const vector<T>& x);
template <class T>
void pr(const set<T>& x);
template <class T1, class T2>
void pr(const map<T1, T2>& x);
template <class T>
void pr(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void pr(const Arg& first, const Args&... rest) {
  pr(first);
  pr(rest...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.first, ", ", x.second, "}");
}
template <class T, bool pretty = true>
void prContain(const T& x) {
  if (pretty) pr("{");
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? pretty ? ", " : " " : "", a), fst = 0;
  if (pretty) pr("}");
}
template <class T>
void pc(const T& x) {
  prContain<T, false>(x);
  pr("\n");
}
template <class T, size_t SZ>
void pr(const array<T, SZ>& x) {
  prContain(x);
}
template <class T>
void pr(const vector<T>& x) {
  prContain(x);
}
template <class T>
void pr(const set<T>& x) {
  prContain(x);
}
template <class T1, class T2>
void pr(const map<T1, T2>& x) {
  prContain(x);
}
void ps() { pr("\n"); }
template <class Arg>
void ps(const Arg& first) {
  pr(first);
  ps();
}
template <class Arg, class... Args>
void ps(const Arg& first, const Args&... rest) {
  pr(first, " ");
  ps(rest...);
}
}  // namespace __output
using namespace __output;
namespace __algorithm {
template <typename T>
void dedup(vector<T>& v) {
  sort((v).begin(), (v).end());
  v.erase(unique((v).begin(), (v).end()), v.end());
}
template <typename T>
typename vector<T>::const_iterator find(const vector<T>& v, const T& x) {
  auto it = lower_bound((v).begin(), (v).end(), x);
  return it != v.end() && *it == x ? it : v.end();
}
template <typename T>
size_t index(const vector<T>& v, const T& x) {
  auto it = find(v, x);
  assert(it != v.end() && *it == x);
  return it - v.begin();
}
template <typename T1, typename T2>
typename vector<pair<T1, T2>>::iterator lower_bound(
    const vector<pair<T1, T2>>& v, const T1& x) {
  return lower_bound(
      (v).begin(), (v).end(), x,
      [](pair<T1, T2> a, pair<T1, T2> b) { return a.first < b.first; });
}
template <typename T1, typename T2>
typename vector<pair<T1, T2>>::iterator upper_bound(
    const vector<pair<T1, T2>>& v, const T1& x) {
  return upper_bound(
      (v).begin(), (v).end(), x,
      [](pair<T1, T2> a, pair<T1, T2> b) { return a.first < b.first; });
}
}  // namespace __algorithm
using namespace __algorithm;
struct __monostate {
  friend istream& operator>>(istream& is,
                             const __attribute__((unused)) __monostate& ms) {
    return is;
  }
  friend ostream& operator<<(ostream& os,
                             const __attribute__((unused)) __monostate& ms) {
    return os;
  }
} ms;
namespace __io {
void setIn(string second) { freopen(second.c_str(), "r", stdin); }
void setOut(string second) { freopen(second.c_str(), "w", stdout); }
void setIO(string second = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << setprecision(15);
  if (int((second).size())) {
    setIn(second + ".in"), setOut(second + ".out");
  }
}
}  // namespace __io
using namespace __io;
void solve() {
  int N, X, Y;
  re(N, X, Y);
  vi b(N);
  re(b);
  for (auto& e : b) --e;
  vvi occur(N + 1);
  for (int i = 0; i < (N); i++) occur[b[i]].push_back(i);
  int unused = 0;
  while (!occur[unused].empty()) unused++;
  auto can_select = [&](int limit) {
    int get = 0;
    for (int c = 0; c < (N + 1); c++) get += min(int((occur[c]).size()), limit);
    return get >= N - X;
  };
  vi chosen, rest;
  auto select = [&](int limit) {
    chosen.clear();
    rest.clear();
    for (int c = 0; c < (N + 1); c++) {
      int take =
          min({N - X - int((chosen).size()), limit, int((occur[c]).size())});
      for (int i = 0; i < (int((occur[c]).size())); i++)
        (i < take ? chosen : rest).push_back(occur[c][i]);
    }
  };
  assert(can_select(N - X));
  select(N - X);
  int lo = 0, hi = N - X;
  while (hi - lo > 1) {
    int mi = lo + (hi - lo) / 2;
    if (can_select(mi)) {
      select(mi);
      hi = mi;
    } else {
      lo = mi;
    }
  }
  vi a(N, unused);
  for (auto& e : rest) a[e] = b[e];
  if (2 * hi > N - X) {
    int spots = N - X - hi;
    if (2 * spots < Y - X) {
      ps("NO");
      return;
    }
    vi count(N + 1);
    for (auto& e : chosen) count[b[e]]++;
    sort((chosen).begin(), (chosen).end(), [&](int i, int j) {
      return make_pair(count[b[i]], b[i]) > make_pair(count[b[j]], b[j]);
    });
    for (int i = 0; i < (Y - X); i++) {
      int use = N - X - 2 * spots + i;
      int put = (use + spots) % int((chosen).size());
      assert(b[chosen[put]] != b[chosen[use]]);
      a[chosen[put]] = b[chosen[use]];
    }
  } else {
    for (int i = 0; i < (Y - X); i++) {
      int j = (i + (N - X) / 2) % int((chosen).size());
      assert(b[chosen[j]] != b[chosen[i]]);
      a[chosen[j]] = b[chosen[i]];
    }
  }
  for (auto& e : a) ++e;
  ps("YES");
  pc(a);
}
int main() {
  setIO();
  int T;
  re(T);
  for (int t = 0; t < (T); t++) {
    solve();
  }
  return 0;
}