#include <bits/stdc++.h>
using namespace std;
using P = pair<int, int>;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n, t, k, x;
  int arr[200200];
  int ans[200200], top;
  cin >> t;
  while (t--) {
    cin >> n;
    x = 0;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      x += arr[i];
    }
    if (x % 2 == 1) {
      cout << "NO\n";
      continue;
    }
    top = 0;
    if (arr[0] == 1 && arr[n - 1] == 1) {
      for (k = 0; arr[k] == 1 && k < n; k++)
        ;
      if (k % 2 == 0) {
        if (k == n) {
          cout << "NO\n";
          continue;
        }
        for (int i = k - 1; i > 0; i -= 2) {
          ans[top++] = i;
          arr[i - 1] = 0;
          arr[i] = 0;
        }
      } else if (k + 1 < n && arr[k + 1] == 1) {
        ans[top++] = k;
        arr[k - 1] = 0;
        arr[k + 1] = 0;
        for (int i = k - 2; i > 0; i -= 2) {
          ans[top++] = i;
          arr[i - 1] = 0;
          arr[i] = 0;
        }
      } else {
        for (int i = k; i < n - 1; i += 2) {
          ans[top++] = i;
          x = arr[i - 1] ^ arr[i] ^ arr[i + 1];
          arr[i - 1] = x;
          arr[i] = x;
          arr[i + 1] = x;
          if (x == 0) {
            for (int j = i - 2; j > 0; j -= 2) {
              ans[top++] = j;
              arr[j] = 0;
              arr[j - 1] = 0;
            }
            break;
          }
        }
      }
    }
    if (arr[0] == 0) {
      for (int i = 1; i < n - 1; i++) {
        if (arr[i] == 1) {
          if (arr[i + 1] == 1) {
            ans[top++] = i;
            arr[i] = 0;
            arr[i + 1] = 0;
          } else if (arr[i + 2] == 1) {
            ans[top++] = i + 1;
            arr[i] = 0;
            arr[i + 2] = 0;
          } else {
            ans[top++] = i + 1;
            ans[top++] = i;
            arr[i] = 0;
            arr[i + 2] = 1;
            i++;
          }
        }
      }
    } else if (arr[n - 1] == 0) {
      for (int i = n - 2; i > 0; i--) {
        if (arr[i] == 1) {
          if (arr[i - 1] == 1) {
            ans[top++] = i;
            arr[i] = 0;
            arr[i - 1] = 0;
          } else if (arr[i - 2] == 1) {
            ans[top++] = i - 1;
            arr[i] = 0;
            arr[i - 2] = 0;
          } else {
            ans[top++] = i - 1;
            ans[top++] = i;
            arr[i] = 0;
            arr[i - 2] = 1;
            i--;
          }
        }
      }
    } else {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    cout << top << '\n';
    for (int i = 0; i < top; i++) {
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
  cout << flush;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 P = pair<int, int>;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n, t, k, x;
  int arr[200200];
  int ans[200200], top;
  cin >> t;
  while (t--) {
    cin >> n;
    x = 0;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      x += arr[i];
    }
    if (x % 2 == 1) {
      cout << "NO\n";
      continue;
    }
    top = 0;
    if (arr[0] == 1 && arr[n - 1] == 1) {
      for (k = 0; arr[k] == 1 && k < n; k++)
        ;
      if (k % 2 == 0) {
        if (k == n) {
          cout << "NO\n";
          continue;
        }
        for (int i = k - 1; i > 0; i -= 2) {
          ans[top++] = i;
          arr[i - 1] = 0;
          arr[i] = 0;
        }
      } else if (k + 1 < n && arr[k + 1] == 1) {
        ans[top++] = k;
        arr[k - 1] = 0;
        arr[k + 1] = 0;
        for (int i = k - 2; i > 0; i -= 2) {
          ans[top++] = i;
          arr[i - 1] = 0;
          arr[i] = 0;
        }
      } else {
        for (int i = k; i < n - 1; i += 2) {
          ans[top++] = i;
          x = arr[i - 1] ^ arr[i] ^ arr[i + 1];
          arr[i - 1] = x;
          arr[i] = x;
          arr[i + 1] = x;
          if (x == 0) {
            for (int j = i - 2; j > 0; j -= 2) {
              ans[top++] = j;
              arr[j] = 0;
              arr[j - 1] = 0;
            }
            break;
          }
        }
      }
    }
    if (arr[0] == 0) {
      for (int i = 1; i < n - 1; i++) {
        if (arr[i] == 1) {
          if (arr[i + 1] == 1) {
            ans[top++] = i;
            arr[i] = 0;
            arr[i + 1] = 0;
          } else if (arr[i + 2] == 1) {
            ans[top++] = i + 1;
            arr[i] = 0;
            arr[i + 2] = 0;
          } else {
            ans[top++] = i + 1;
            ans[top++] = i;
            arr[i] = 0;
            arr[i + 2] = 1;
            i++;
          }
        }
      }
    } else if (arr[n - 1] == 0) {
      for (int i = n - 2; i > 0; i--) {
        if (arr[i] == 1) {
          if (arr[i - 1] == 1) {
            ans[top++] = i;
            arr[i] = 0;
            arr[i - 1] = 0;
          } else if (arr[i - 2] == 1) {
            ans[top++] = i - 1;
            arr[i] = 0;
            arr[i - 2] = 0;
          } else {
            ans[top++] = i - 1;
            ans[top++] = i;
            arr[i] = 0;
            arr[i - 2] = 1;
            i--;
          }
        }
      }
    } else {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    cout << top << '\n';
    for (int i = 0; i < top; i++) {
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
  cout << flush;
  return 0;
}

```