#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int times;
  cin >> times;
  while (times--) {
    int n;
    cin >> n;
    vector<int> x(n);
    for (int i = 0; i < n; ++i) cin >> x[i];
    vector<int> ans;
    bool flag = true;
    for (int i = 0; i < n;) {
      if (x[i] == 0) {
        ++i;
        continue;
      }
      int j = i;
      while (j < n && x[j] == 1) ++j;
      while ((j - i) & 1) {
        int oj = j;
        while (j + 1 < n && x[j] == 0 && x[j + 1] == 0) {
          ans.push_back(j);
          x[j] = x[j + 1] = 1;
          j += 2;
        }
        while (j < n && x[j] == 1) ++j;
        if (oj == j) break;
      }
      if ((j - i) & 1) {
        if (j == n || j + 1 == n) {
          flag = false;
          break;
        }
        x[j + 1] = 0;
        for (int k = j - 1; k >= i; k -= 2) {
          ans.push_back(k + 1);
        }
      } else {
        if (i > 0) {
          for (int k = i; k < j; k += 2) {
            ans.push_back(k);
          }
        } else if (j < n) {
          for (int k = j - 2; k >= i; k -= 2) {
            ans.push_back(k + 1);
          }
        } else {
          flag = false;
          break;
        }
      }
      i = j;
    }
    if (flag) {
      cout << "YES\n";
      cout << ans.size() << "\n";
      for (int i : ans) cout << i << " ";
      cout << "\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}

### 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;
using LL = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int times;
  cin >> times;
  while (times--) {
    int n;
    cin >> n;
    vector<int> x(n);
    for (int i = 0; i < n; ++i) cin >> x[i];
    vector<int> ans;
    bool flag = true;
    for (int i = 0; i < n;) {
      if (x[i] == 0) {
        ++i;
        continue;
      }
      int j = i;
      while (j < n && x[j] == 1) ++j;
      while ((j - i) & 1) {
        int oj = j;
        while (j + 1 < n && x[j] == 0 && x[j + 1] == 0) {
          ans.push_back(j);
          x[j] = x[j + 1] = 1;
          j += 2;
        }
        while (j < n && x[j] == 1) ++j;
        if (oj == j) break;
      }
      if ((j - i) & 1) {
        if (j == n || j + 1 == n) {
          flag = false;
          break;
        }
        x[j + 1] = 0;
        for (int k = j - 1; k >= i; k -= 2) {
          ans.push_back(k + 1);
        }
      } else {
        if (i > 0) {
          for (int k = i; k < j; k += 2) {
            ans.push_back(k);
          }
        } else if (j < n) {
          for (int k = j - 2; k >= i; k -= 2) {
            ans.push_back(k + 1);
          }
        } else {
          flag = false;
          break;
        }
      }
      i = j;
    }
    if (flag) {
      cout << "YES\n";
      cout << ans.size() << "\n";
      for (int i : ans) cout << i << " ";
      cout << "\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}

```