#include <bits/stdc++.h>
using namespace std;
void testcase() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int x = 0;
  for (int i = 0; i < n; i++) {
    x = x ^ a[i];
  }
  if (x != 0) {
    cout << "NO" << '\n';
    return;
  }
  if (n % 2 == 1) {
    cout << "YES" << '\n';
    if (n == 3) {
      cout << "1" << '\n' << "1" << '\n';
      return;
    }
    cout << n - 2 << '\n';
    for (int i = 1; i <= n - 2; i += 2) {
      cout << i << " ";
    }
    for (int i = n - 4; i >= 1; i -= 2) {
      cout << i;
      if (i == 1) {
        cout << '\n';
      } else {
        cout << " ";
      }
    }
  } else {
    int y = a[0];
    int k = 1;
    while ((y != 0) && (k < n - 1)) {
      y = y ^ a[k];
      y = y ^ a[k + 1];
      k += 2;
    }
    if (y == 1) {
      cout << "NO" << '\n';
      return;
    } else {
      cout << "YES" << '\n';
      if (k == 1) {
        cout << n - 3 << '\n';
        if (n == 4) {
          cout << "2" << '\n';
          return;
        }
        for (int i = 2; i <= n - 2; i += 2) {
          cout << i << " ";
        }
        for (int i = n - 4; i >= 2; i -= 2) {
          cout << i;
          if (i == 2) {
            cout << '\n';
          } else {
            cout << " ";
          }
        }
      } else if (k == n - 1) {
        cout << n - 3 << '\n';
        if (n == 4) {
          cout << "1" << '\n';
          return;
        }
        for (int i = 1; i <= n - 3; i += 2) {
          cout << i << " ";
        }
        for (int i = n - 5; i >= 1; i -= 2) {
          cout << i;
          if (i == 1) {
            cout << '\n';
          } else {
            cout << " ";
          }
        }
      } else {
        cout << n - 4 << '\n';
        for (int i = 1; i <= k - 2; i += 2) {
          cout << i << " ";
        }
        for (int i = k - 4; i >= 1; i -= 2) {
          cout << i << " ";
        }
        if (n - k == 3) {
          cout << k + 1 << '\n';
          return;
        }
        for (int i = 1; i <= n - k - 2; i += 2) {
          cout << k + i << " ";
        }
        for (int i = n - k - 4; i >= 1; i -= 2) {
          cout << i + k;
          if (i == 1) {
            cout << '\n';
          } else {
            cout << " ";
          }
        }
      }
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    testcase();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 testcase() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int x = 0;
  for (int i = 0; i < n; i++) {
    x = x ^ a[i];
  }
  if (x != 0) {
    cout << "NO" << '\n';
    return;
  }
  if (n % 2 == 1) {
    cout << "YES" << '\n';
    if (n == 3) {
      cout << "1" << '\n' << "1" << '\n';
      return;
    }
    cout << n - 2 << '\n';
    for (int i = 1; i <= n - 2; i += 2) {
      cout << i << " ";
    }
    for (int i = n - 4; i >= 1; i -= 2) {
      cout << i;
      if (i == 1) {
        cout << '\n';
      } else {
        cout << " ";
      }
    }
  } else {
    int y = a[0];
    int k = 1;
    while ((y != 0) && (k < n - 1)) {
      y = y ^ a[k];
      y = y ^ a[k + 1];
      k += 2;
    }
    if (y == 1) {
      cout << "NO" << '\n';
      return;
    } else {
      cout << "YES" << '\n';
      if (k == 1) {
        cout << n - 3 << '\n';
        if (n == 4) {
          cout << "2" << '\n';
          return;
        }
        for (int i = 2; i <= n - 2; i += 2) {
          cout << i << " ";
        }
        for (int i = n - 4; i >= 2; i -= 2) {
          cout << i;
          if (i == 2) {
            cout << '\n';
          } else {
            cout << " ";
          }
        }
      } else if (k == n - 1) {
        cout << n - 3 << '\n';
        if (n == 4) {
          cout << "1" << '\n';
          return;
        }
        for (int i = 1; i <= n - 3; i += 2) {
          cout << i << " ";
        }
        for (int i = n - 5; i >= 1; i -= 2) {
          cout << i;
          if (i == 1) {
            cout << '\n';
          } else {
            cout << " ";
          }
        }
      } else {
        cout << n - 4 << '\n';
        for (int i = 1; i <= k - 2; i += 2) {
          cout << i << " ";
        }
        for (int i = k - 4; i >= 1; i -= 2) {
          cout << i << " ";
        }
        if (n - k == 3) {
          cout << k + 1 << '\n';
          return;
        }
        for (int i = 1; i <= n - k - 2; i += 2) {
          cout << k + i << " ";
        }
        for (int i = n - k - 4; i >= 1; i -= 2) {
          cout << i + k;
          if (i == 1) {
            cout << '\n';
          } else {
            cout << " ";
          }
        }
      }
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    testcase();
  }
  return 0;
}

```