#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 10;
long long n;
long long a[maxn];
void getInput() {
  cin >> n;
  for (long long i = (1); i <= (n); i++) cin >> a[i];
}
void solve() {
  getInput();
  vector<long long> ans;
  long long ones = 0;
  for (long long i = (1); i <= (n); i++) ones += a[i];
  if (ones & 1) {
    cout << "NO\n";
    return;
  }
  if (n & 1) {
    for (long long i = n - 2; i >= 1; i -= 2) ans.push_back(i);
    for (long long i = 1; i <= n - 2; i += 2) ans.push_back(i);
    cout << "YES\n";
    cout << ((long long)ans.size()) << '\n';
    for (auto x : ans) cout << x << " ";
    cout << '\n';
    return;
  } else {
    ones = 0;
    for (long long i = 1; i <= n; i += 2) {
      ones += a[i - 1] + a[i];
      if (ones % 2 == 0) {
        long long k = i;
        for (long long j = k - 2; j >= 1; j -= 2) ans.push_back(j);
        for (long long j = 1; j <= k - 2; j += 2) ans.push_back(j);
        for (long long j = n - 2; j >= k + 1; j -= 2) ans.push_back(j);
        for (long long j = k + 1; j <= n - 2; j += 2) ans.push_back(j);
        cout << "YES\n";
        cout << ((long long)ans.size()) << '\n';
        for (auto x : ans) cout << x << " ";
        cout << '\n';
        return;
      }
    }
    cout << "NO\n";
    return;
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long T;
  cin >> T;
  while (T--) solve();
}

### Prompt

Create a solution in cpp for 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;
const long long maxn = 2e5 + 10;
long long n;
long long a[maxn];
void getInput() {
  cin >> n;
  for (long long i = (1); i <= (n); i++) cin >> a[i];
}
void solve() {
  getInput();
  vector<long long> ans;
  long long ones = 0;
  for (long long i = (1); i <= (n); i++) ones += a[i];
  if (ones & 1) {
    cout << "NO\n";
    return;
  }
  if (n & 1) {
    for (long long i = n - 2; i >= 1; i -= 2) ans.push_back(i);
    for (long long i = 1; i <= n - 2; i += 2) ans.push_back(i);
    cout << "YES\n";
    cout << ((long long)ans.size()) << '\n';
    for (auto x : ans) cout << x << " ";
    cout << '\n';
    return;
  } else {
    ones = 0;
    for (long long i = 1; i <= n; i += 2) {
      ones += a[i - 1] + a[i];
      if (ones % 2 == 0) {
        long long k = i;
        for (long long j = k - 2; j >= 1; j -= 2) ans.push_back(j);
        for (long long j = 1; j <= k - 2; j += 2) ans.push_back(j);
        for (long long j = n - 2; j >= k + 1; j -= 2) ans.push_back(j);
        for (long long j = k + 1; j <= n - 2; j += 2) ans.push_back(j);
        cout << "YES\n";
        cout << ((long long)ans.size()) << '\n';
        for (auto x : ans) cout << x << " ";
        cout << '\n';
        return;
      }
    }
    cout << "NO\n";
    return;
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long T;
  cin >> T;
  while (T--) solve();
}

```