#include <bits/stdc++.h>
using namespace std;
int s[1000001];
vector<int> q;
void add(int l, int r) {
  for (int x = r - 2; x >= l; x -= 2) q.push_back(x);
  for (int x = l; x < r; x += 2) q.push_back(x);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int x = 1, y; x <= n; x++) cin >> y, s[x] = s[x - 1] ^ y;
    if (!s[n])
      if (n & 1)
        add(1, n);
      else
        for (int x = 1; x <= n; x += 2)
          if (!s[x]) {
            add(1, x);
            add(x + 1, n);
            break;
          }
    if (q.size())
      cout << "YES" << endl << q.size() << endl;
    else
      cout << "NO";
    for (auto x : q) cout << x << ' ';
    vector<int>().swap(q);
    cout << endl;
  }
}

### 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;
int s[1000001];
vector<int> q;
void add(int l, int r) {
  for (int x = r - 2; x >= l; x -= 2) q.push_back(x);
  for (int x = l; x < r; x += 2) q.push_back(x);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int x = 1, y; x <= n; x++) cin >> y, s[x] = s[x - 1] ^ y;
    if (!s[n])
      if (n & 1)
        add(1, n);
      else
        for (int x = 1; x <= n; x += 2)
          if (!s[x]) {
            add(1, x);
            add(x + 1, n);
            break;
          }
    if (q.size())
      cout << "YES" << endl << q.size() << endl;
    else
      cout << "NO";
    for (auto x : q) cout << x << ' ';
    vector<int>().swap(q);
    cout << endl;
  }
}

```