#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
template <class T>
int size(T &&a) {
  return (int)(a.size());
}
ostream &operator<<(ostream &os, string str) {
  for (char c : str) os << c;
  return os;
}
template <class A, class B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
template <class T>
auto operator<<(ostream &os, T &&x) -> decltype(x.begin(), os) {
  os << '{';
  for (auto it = x.begin(); it != x.end(); ++it)
    os << *it << (it == prev(x.end()) ? "" : " ");
  return os << '}';
}
template <class T>
ostream &operator<<(ostream &os, vector<vector<T>> vec) {
  for (auto x : vec) os << "\n  " << x;
  return os;
}
void dump() {}
template <class T, class... Args>
void dump(T &&x, Args... args) {
  cerr << x << "; ";
  dump(args...);
}
vector<int> ans;
void print_sequence(int st, int en) {
  if (en - st + 1 < 3) return;
  for (int i = st; i <= en - 2; i += 2) ans.emplace_back(i);
  for (int i = en - 4; i >= st; i -= 2) ans.emplace_back(i);
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n), pref_xor(n);
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    if (i == 0)
      pref_xor[i] = a[i];
    else
      pref_xor[i] = pref_xor[i - 1] ^ a[i];
  }
  if (pref_xor[n - 1] == 1) {
    cout << "NO\n";
    return;
  }
  ans.clear();
  if (n % 2 == 1) {
    cout << "YES\n";
    print_sequence(0, n - 1);
    cout << size(ans) << '\n';
    for (int x : ans) cout << x + 1 << ' ';
    cout << '\n';
    return;
  }
  for (int i = 0; i < n; i += 2) {
    if (pref_xor[i] == 0) {
      cout << "YES\n";
      print_sequence(0, i);
      print_sequence(i + 1, n - 1);
      cout << size(ans) << '\n';
      for (int x : ans) cout << x + 1 << ' ';
      cout << '\n';
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
}

### Prompt

Generate a cpp solution to the following problem:

You are given a sequence a of length n consisting of 0s and 1s.

You can perform the following operation on this sequence: 

  * Pick an index i from 1 to n-2 (inclusive). 
  * Change all of a_{i}, a_{i+1}, a_{i+2} to a_{i} ⊕ a_{i+1} ⊕ a_{i+2} simultaneously, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)

Find a sequence of at most n operations that changes all elements of a to 0s or report that it's impossible.

We can prove that if there exists a sequence of operations of any length that changes all elements of a to 0s, then there is also such a sequence of length not greater than n.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4).

The first line of each test case contains a single integer n (3 ≤ n ≤ 2⋅10^5) — the length of a.

The second line of each test case contains n integers a_1, a_2, …, a_n (a_i = 0 or a_i = 1) — elements of a.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5.

Output

For each test case, do the following: 

  * if there is no way of making all the elements of a equal to 0 after performing the above operation some number of times, print "NO". 
  * otherwise, in the first line print "YES", in the second line print k (0 ≤ k ≤ n) — the number of operations that you want to perform on a, and in the third line print a sequence b_1, b_2, ..., b_k (1 ≤ b_i ≤ n - 2) — the indices on which the operation should be applied. 



If there are multiple solutions, you may print any.

Example

Input


3
3
0 0 0
5
1 1 1 1 0
4
1 0 0 1


Output


YES
0
YES
2
3 1
NO

Note

In the first example, the sequence contains only 0s so we don't need to change anything.

In the second example, we can transform [1, 1, 1, 1, 0] to [1, 1, 0, 0, 0] and then to [0, 0, 0, 0, 0] by performing the operation on the third element of a and then on the first element of a.

In the third example, no matter whether we first perform the operation on the first or on the second element of a we will get [1, 1, 1, 1], which cannot be transformed to [0, 0, 0, 0].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
template <class T>
int size(T &&a) {
  return (int)(a.size());
}
ostream &operator<<(ostream &os, string str) {
  for (char c : str) os << c;
  return os;
}
template <class A, class B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
template <class T>
auto operator<<(ostream &os, T &&x) -> decltype(x.begin(), os) {
  os << '{';
  for (auto it = x.begin(); it != x.end(); ++it)
    os << *it << (it == prev(x.end()) ? "" : " ");
  return os << '}';
}
template <class T>
ostream &operator<<(ostream &os, vector<vector<T>> vec) {
  for (auto x : vec) os << "\n  " << x;
  return os;
}
void dump() {}
template <class T, class... Args>
void dump(T &&x, Args... args) {
  cerr << x << "; ";
  dump(args...);
}
vector<int> ans;
void print_sequence(int st, int en) {
  if (en - st + 1 < 3) return;
  for (int i = st; i <= en - 2; i += 2) ans.emplace_back(i);
  for (int i = en - 4; i >= st; i -= 2) ans.emplace_back(i);
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n), pref_xor(n);
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    if (i == 0)
      pref_xor[i] = a[i];
    else
      pref_xor[i] = pref_xor[i - 1] ^ a[i];
  }
  if (pref_xor[n - 1] == 1) {
    cout << "NO\n";
    return;
  }
  ans.clear();
  if (n % 2 == 1) {
    cout << "YES\n";
    print_sequence(0, n - 1);
    cout << size(ans) << '\n';
    for (int x : ans) cout << x + 1 << ' ';
    cout << '\n';
    return;
  }
  for (int i = 0; i < n; i += 2) {
    if (pref_xor[i] == 0) {
      cout << "YES\n";
      print_sequence(0, i);
      print_sequence(i + 1, n - 1);
      cout << size(ans) << '\n';
      for (int x : ans) cout << x + 1 << ' ';
      cout << '\n';
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
}

```