#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& x : a) cin >> x;
  int asum = 0;
  for (int x : a) asum ^= x;
  if (asum) {
    cout << "NO" << '\n';
    return;
  }
  vector<int> b(n + 1);
  b[0] = a[0];
  for (int i = 0; i < n - 1; i++) {
    b[i + 1] = a[i + 1] ^ a[i];
  }
  b[n] = a[n - 1];
  vector<int> steps;
  if (n & 1) {
    for (int i = 0; i <= n - 3; i += 2) steps.push_back(i);
    for (int i = n - 3; i >= 0; i -= 2) steps.push_back(i);
  } else {
    int split = -1;
    int psum = 0;
    for (int i = 0; i < n; i += 2) {
      psum ^= b[i];
      if (psum == 0) {
        split = i;
        break;
      }
    }
    if (split == -1) {
      cout << "NO" << '\n';
      return;
    }
    for (int i = split - 2; i >= 0; i -= 2) {
      steps.push_back(i);
    }
    for (int i = split + 2; i < n; i += 2) {
      steps.push_back(i - 1);
    }
    for (int i = 0; i < n - 2; i += 2) {
      steps.push_back(i);
    }
  }
  cout << "YES" << '\n';
  cout << steps.size() << '\n';
  for (int x : steps) {
    cout << (x + 1) << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) solve();
}

### Prompt

Your challenge is to write 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;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& x : a) cin >> x;
  int asum = 0;
  for (int x : a) asum ^= x;
  if (asum) {
    cout << "NO" << '\n';
    return;
  }
  vector<int> b(n + 1);
  b[0] = a[0];
  for (int i = 0; i < n - 1; i++) {
    b[i + 1] = a[i + 1] ^ a[i];
  }
  b[n] = a[n - 1];
  vector<int> steps;
  if (n & 1) {
    for (int i = 0; i <= n - 3; i += 2) steps.push_back(i);
    for (int i = n - 3; i >= 0; i -= 2) steps.push_back(i);
  } else {
    int split = -1;
    int psum = 0;
    for (int i = 0; i < n; i += 2) {
      psum ^= b[i];
      if (psum == 0) {
        split = i;
        break;
      }
    }
    if (split == -1) {
      cout << "NO" << '\n';
      return;
    }
    for (int i = split - 2; i >= 0; i -= 2) {
      steps.push_back(i);
    }
    for (int i = split + 2; i < n; i += 2) {
      steps.push_back(i - 1);
    }
    for (int i = 0; i < n - 2; i += 2) {
      steps.push_back(i);
    }
  }
  cout << "YES" << '\n';
  cout << steps.size() << '\n';
  for (int x : steps) {
    cout << (x + 1) << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) solve();
}

```