#include <bits/stdc++.h>
const int N = 2000500;
using namespace std;
int pr[2], f[N], n, t, a[N], s[N];
queue<int> q;
void out(int l, int r) {
  for (int i = r - 2; i >= l; i -= 2) q.push(i);
  for (int i = l; i < r; i += 2) q.push(i);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i], s[i] = a[i] ^ s[i - 1];
    if (!s[n]) {
      if (n & 1)
        out(1, n);
      else {
        for (int i = 1; i <= n; i += 2) {
          if (s[i] == 0) {
            out(1, i);
            out(i + 1, n);
            break;
          }
        }
      }
    }
    if (q.size())
      cout << "YES" << endl << q.size() << endl;
    else
      cout << "NO";
    while (q.size()) cout << q.front() << ' ', q.pop();
    cout << endl;
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
const int N = 2000500;
using namespace std;
int pr[2], f[N], n, t, a[N], s[N];
queue<int> q;
void out(int l, int r) {
  for (int i = r - 2; i >= l; i -= 2) q.push(i);
  for (int i = l; i < r; i += 2) q.push(i);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i], s[i] = a[i] ^ s[i - 1];
    if (!s[n]) {
      if (n & 1)
        out(1, n);
      else {
        for (int i = 1; i <= n; i += 2) {
          if (s[i] == 0) {
            out(1, i);
            out(i + 1, n);
            break;
          }
        }
      }
    }
    if (q.size())
      cout << "YES" << endl << q.size() << endl;
    else
      cout << "NO";
    while (q.size()) cout << q.front() << ' ', q.pop();
    cout << endl;
  }
}

```