#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  for (; t--;) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int &i : a) {
      cin >> i;
    }
    auto oper = [&](int i) {
      int x = 0;
      for (int j = 0; j < 3; j++) x ^= a[i + j];
      for (int j = 0; j < 3; j++) a[i + j] = x;
    };
    vector<int> ans;
    for (int i = 0; i < n - 2;) {
      int cnt = 0;
      for (int j = 0; j < 3; j++) cnt += a[i + j];
      if (cnt & 1) {
        if (cnt == 1 && !a[i]) {
          i++;
        } else {
          if (cnt == 1) {
            oper(i);
            ans.push_back(i + 1);
          }
          i += 2;
        }
      } else {
        i += 3;
      }
    }
    for (int i = n - 3; i >= 0; i--) {
      int cnt = 0;
      for (int j = 0; j < 3; j++) cnt += a[i + j];
      if (cnt == 2) {
        ans.push_back(i + 1);
        oper(i);
      } else if (cnt == 3) {
        i--;
      }
    }
    for (int i = 0; i < n - 2; i++) {
      int cnt = 0;
      for (int j = 0; j < 3; j++) cnt += a[i + j];
      if (cnt == 2) {
        ans.push_back(i + 1);
        oper(i);
      }
    }
    bool ok = 1;
    for (int i : a)
      if (i) ok = 0;
    if (ok) {
      cout << "YES\n";
      cout << ans.size() << '\n';
      for (int i : ans) cout << i << ' ';
      cout << '\n';
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  for (; t--;) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int &i : a) {
      cin >> i;
    }
    auto oper = [&](int i) {
      int x = 0;
      for (int j = 0; j < 3; j++) x ^= a[i + j];
      for (int j = 0; j < 3; j++) a[i + j] = x;
    };
    vector<int> ans;
    for (int i = 0; i < n - 2;) {
      int cnt = 0;
      for (int j = 0; j < 3; j++) cnt += a[i + j];
      if (cnt & 1) {
        if (cnt == 1 && !a[i]) {
          i++;
        } else {
          if (cnt == 1) {
            oper(i);
            ans.push_back(i + 1);
          }
          i += 2;
        }
      } else {
        i += 3;
      }
    }
    for (int i = n - 3; i >= 0; i--) {
      int cnt = 0;
      for (int j = 0; j < 3; j++) cnt += a[i + j];
      if (cnt == 2) {
        ans.push_back(i + 1);
        oper(i);
      } else if (cnt == 3) {
        i--;
      }
    }
    for (int i = 0; i < n - 2; i++) {
      int cnt = 0;
      for (int j = 0; j < 3; j++) cnt += a[i + j];
      if (cnt == 2) {
        ans.push_back(i + 1);
        oper(i);
      }
    }
    bool ok = 1;
    for (int i : a)
      if (i) ok = 0;
    if (ok) {
      cout << "YES\n";
      cout << ans.size() << '\n';
      for (int i : ans) cout << i << ' ';
      cout << '\n';
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}

```