#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<int> res;
    int p = 0;
    while (p + 2 < n) {
      if (a[p] == 0 && a[p + 1] == 1 && a[p + 2] == 1) {
        res.push_back(p);
        a[p + 1] = 0;
        a[p + 2] = 0;
        p += 2;
        continue;
      } else if (a[p] == 1) {
        if (a[p + 1] == 0 && a[p + 2] == 0) {
          res.push_back(p);
          a[p + 1] = 1;
          a[p + 2] = 1;
        } else if (a[p + 1] == 0 || a[p + 2] == 0) {
          res.push_back(p);
          a[p + 0] = 0;
          a[p + 1] = 0;
          a[p + 2] = 0;
          int pp = p - 2;
          while (pp >= 0 && a[pp] == 1) {
            res.push_back(pp);
            a[pp] = 0;
            a[pp + 1] = 0;
            pp -= 2;
          }
        }
        p += 2;
        continue;
      }
      p++;
    }
    int x = n - 1;
    while (x >= 0 && a[x] == 1) x--;
    if (x >= 0) {
      while (x + 2 < n) {
        res.push_back(x);
        a[x + 1] = 0;
        a[x + 2] = 0;
        x += 2;
      }
    }
    x = n - 1;
    while (x >= 0 && a[x] == 0) x--;
    if (x >= 0) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      cout << res.size() << endl;
      if (res.size() > 0) {
        for (int v : res) {
          cout << v + 1 << " ";
        }
        cout << "\n";
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<int> res;
    int p = 0;
    while (p + 2 < n) {
      if (a[p] == 0 && a[p + 1] == 1 && a[p + 2] == 1) {
        res.push_back(p);
        a[p + 1] = 0;
        a[p + 2] = 0;
        p += 2;
        continue;
      } else if (a[p] == 1) {
        if (a[p + 1] == 0 && a[p + 2] == 0) {
          res.push_back(p);
          a[p + 1] = 1;
          a[p + 2] = 1;
        } else if (a[p + 1] == 0 || a[p + 2] == 0) {
          res.push_back(p);
          a[p + 0] = 0;
          a[p + 1] = 0;
          a[p + 2] = 0;
          int pp = p - 2;
          while (pp >= 0 && a[pp] == 1) {
            res.push_back(pp);
            a[pp] = 0;
            a[pp + 1] = 0;
            pp -= 2;
          }
        }
        p += 2;
        continue;
      }
      p++;
    }
    int x = n - 1;
    while (x >= 0 && a[x] == 1) x--;
    if (x >= 0) {
      while (x + 2 < n) {
        res.push_back(x);
        a[x + 1] = 0;
        a[x + 2] = 0;
        x += 2;
      }
    }
    x = n - 1;
    while (x >= 0 && a[x] == 0) x--;
    if (x >= 0) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      cout << res.size() << endl;
      if (res.size() > 0) {
        for (int v : res) {
          cout << v + 1 << " ";
        }
        cout << "\n";
      }
    }
  }
  return 0;
}

```