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].
#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, su = 0;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    su += a[i];
  }
  if (su % 2) {
    cout << "NO\n";
    return;
  }
  long long s = 0, ind = -1;
  for (long long i = 0; i < n; ++i) {
    if (!a[i] && s % 2 == 0) {
      ind = i;
      break;
    }
    s += a[i];
  }
  vector<long long> ans;
  if (ind == -1) {
    for (long long i = n - 2; i > 0; i -= 2) {
      if (a[i - 1] && a[i]) continue;
      ans.push_back(i);
      a[i] = 1;
      if (a[i - 1]) {
        ind = i - 1;
        a[i - 1] = a[i] = a[i + 1] = 0;
        break;
      }
      a[i - 1] = 1;
    }
    if (ind == -1) {
      cout << "NO\n";
      return;
    }
  }
  for (long long i = ind - 1; i >= 0; --i) {
    if (a[i]) {
      if (a[i - 1]) {
        ans.push_back(i);
        a[i - 1] = a[i] = 0;
      } else {
        if (a[i - 2]) {
          a.push_back(i - 1);
          a[i - 2] = a[i] = 0;
        } else {
          a.push_back(i - 1);
          a.push_back(i);
          a[i - 2] = 1;
          a[i] = 0;
        }
      }
    }
  }
  for (long long i = ind + 1; i < n; ++i) {
    if (a[i]) {
      if (a[i + 1]) {
        ans.push_back(i);
        a[i + 1] = a[i] = 0;
      } else {
        if (a[i + 2]) {
          a.push_back(i + 1);
          a[i + 2] = a[i] = 0;
        } else {
          a.push_back(i + 1);
          a.push_back(i);
          a[i + 2] = 1;
          a[i] = 0;
        }
      }
    }
  }
  cout << "YES\n" << ans.size() << '\n';
  for (long long i = 0; i < ans.size(); ++i) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}
