#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n, a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    int cnt1 = 0, cnt0 = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i])
        cnt1++;
      else
        cnt0++;
    }
    if (cnt1 % 2 != 0 || !cnt0)
      cout << "NO" << endl;
    else if (!cnt1)
      cout << "YES\n0" << endl;
    else {
      vector<int> ans;
      int p = 1, fucked = 0;
      while (1) {
        while (p <= n && a[p]) p++;
        if (p > n) {
          fucked = 1;
          break;
        }
        if (p % 2 == 1)
          break;
        else {
          bool ok = 1;
          while (!a[p + 1]) {
            ans.push_back(p - 1);
            a[p] = a[p + 1] = 1;
            p += 2;
            if (a[p]) {
              ok = 0;
              break;
            }
          }
          if (ok) {
            ans.push_back(p - 1);
            a[p - 1] = a[p + 1] = 0;
            p--;
            break;
          }
        }
      }
      if (fucked) {
        cout << "NO" << endl;
        continue;
      }
      int q = p;
      while (p - 2 > 0) {
        ans.push_back(p - 2);
        a[p - 2] = a[p - 1] = 0;
        p -= 2;
      }
      while (1) {
        while (q <= n && !a[q]) q++;
        if (q > n) break;
        p = q;
        while (1) {
          while (p <= n && a[p]) p++;
          if ((p - q) % 2 == 0)
            break;
          else {
            bool ok = 1;
            while (!a[p + 1]) {
              ans.push_back(p - 1);
              a[p] = a[p + 1] = 1;
              p += 2;
              if (a[p]) {
                ok = 0;
                break;
              }
            }
            if (ok) {
              ans.push_back(p - 1);
              a[p - 1] = a[p + 1] = 0;
              p--;
              break;
            }
          }
        }
        while (q < p) {
          ans.push_back(q - 1);
          q += 2;
        }
      }
      cout << "YES\n" << ans.size() << endl;
      for (int i : ans) cout << i << ' ';
      cout << endl;
    }
  }
}

### 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;
const int N = 200005;
int n, a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    int cnt1 = 0, cnt0 = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i])
        cnt1++;
      else
        cnt0++;
    }
    if (cnt1 % 2 != 0 || !cnt0)
      cout << "NO" << endl;
    else if (!cnt1)
      cout << "YES\n0" << endl;
    else {
      vector<int> ans;
      int p = 1, fucked = 0;
      while (1) {
        while (p <= n && a[p]) p++;
        if (p > n) {
          fucked = 1;
          break;
        }
        if (p % 2 == 1)
          break;
        else {
          bool ok = 1;
          while (!a[p + 1]) {
            ans.push_back(p - 1);
            a[p] = a[p + 1] = 1;
            p += 2;
            if (a[p]) {
              ok = 0;
              break;
            }
          }
          if (ok) {
            ans.push_back(p - 1);
            a[p - 1] = a[p + 1] = 0;
            p--;
            break;
          }
        }
      }
      if (fucked) {
        cout << "NO" << endl;
        continue;
      }
      int q = p;
      while (p - 2 > 0) {
        ans.push_back(p - 2);
        a[p - 2] = a[p - 1] = 0;
        p -= 2;
      }
      while (1) {
        while (q <= n && !a[q]) q++;
        if (q > n) break;
        p = q;
        while (1) {
          while (p <= n && a[p]) p++;
          if ((p - q) % 2 == 0)
            break;
          else {
            bool ok = 1;
            while (!a[p + 1]) {
              ans.push_back(p - 1);
              a[p] = a[p + 1] = 1;
              p += 2;
              if (a[p]) {
                ok = 0;
                break;
              }
            }
            if (ok) {
              ans.push_back(p - 1);
              a[p - 1] = a[p + 1] = 0;
              p--;
              break;
            }
          }
        }
        while (q < p) {
          ans.push_back(q - 1);
          q += 2;
        }
      }
      cout << "YES\n" << ans.size() << endl;
      for (int i : ans) cout << i << ' ';
      cout << endl;
    }
  }
}

```