#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int a[N];
vector<int> ans;
void work(int l, int r) {
  if (a[l]) {
    assert(a[r] == 0);
    int t = ans.size();
    reverse(a + l, a + r + 1);
    work(l, r);
    for (int i = t; i < ans.size(); ++i) {
      int p = ans[i] - l + 1;
      p = p + 2;
      p = r - p + 1;
      ans[i] = l + p - 1;
    }
    return;
  }
  while (l < r) {
    if (a[l + 1] == 0) {
      ++l;
      continue;
    }
    assert(l + 1 != r);
    if (a[l + 2]) {
      ans.push_back(l);
      a[l + 1] = a[l + 2] = 0;
      continue;
    }
    assert(l + 2 != r);
    if (a[l + 3]) {
      ans.push_back(l + 1);
      a[l + 1] = a[l + 3] = 0;
      continue;
    }
    ans.push_back(l + 1);
    a[l + 2] = a[l + 3] = 1;
  }
}
int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    int sum = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      sum ^= a[i];
    }
    if (sum % 2) {
      puts("NO");
      continue;
    }
    ans.clear();
    if (a[1] == 0 || a[n] == 0) {
      work(1, n);
    } else {
      int p_sum = 0, i = 1;
      for (i = 1; i <= n; ++i)
        if (a[i])
          p_sum ^= 1;
        else if (p_sum == 0) {
          work(1, i);
          work(i, n);
          break;
        }
      if (i > n) {
        for (i = 1; i <= n; ++i)
          if (a[i] == 0) {
            int j = i;
            while (j <= n && a[j] == 0) ++j;
            --j;
            if ((j - i + 1) % 2) {
              while (j > i) {
                ans.push_back(j - 1);
                a[j - 1] = a[j] = 1;
                j -= 2;
              }
              ans.push_back(i - 1);
              a[i - 1] = a[i + 1] = 0;
              work(1, i);
              work(i, n);
              break;
            }
            i = j + 1;
          }
        if (i > n) {
          puts("NO");
          continue;
        }
      }
    }
    puts("YES");
    cout << ans.size() << endl;
    for (auto x : ans) printf("%d ", x);
    puts("");
  }
}

### Prompt

Create a solution in Cpp for 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 = 2e5 + 5;
int a[N];
vector<int> ans;
void work(int l, int r) {
  if (a[l]) {
    assert(a[r] == 0);
    int t = ans.size();
    reverse(a + l, a + r + 1);
    work(l, r);
    for (int i = t; i < ans.size(); ++i) {
      int p = ans[i] - l + 1;
      p = p + 2;
      p = r - p + 1;
      ans[i] = l + p - 1;
    }
    return;
  }
  while (l < r) {
    if (a[l + 1] == 0) {
      ++l;
      continue;
    }
    assert(l + 1 != r);
    if (a[l + 2]) {
      ans.push_back(l);
      a[l + 1] = a[l + 2] = 0;
      continue;
    }
    assert(l + 2 != r);
    if (a[l + 3]) {
      ans.push_back(l + 1);
      a[l + 1] = a[l + 3] = 0;
      continue;
    }
    ans.push_back(l + 1);
    a[l + 2] = a[l + 3] = 1;
  }
}
int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    int sum = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      sum ^= a[i];
    }
    if (sum % 2) {
      puts("NO");
      continue;
    }
    ans.clear();
    if (a[1] == 0 || a[n] == 0) {
      work(1, n);
    } else {
      int p_sum = 0, i = 1;
      for (i = 1; i <= n; ++i)
        if (a[i])
          p_sum ^= 1;
        else if (p_sum == 0) {
          work(1, i);
          work(i, n);
          break;
        }
      if (i > n) {
        for (i = 1; i <= n; ++i)
          if (a[i] == 0) {
            int j = i;
            while (j <= n && a[j] == 0) ++j;
            --j;
            if ((j - i + 1) % 2) {
              while (j > i) {
                ans.push_back(j - 1);
                a[j - 1] = a[j] = 1;
                j -= 2;
              }
              ans.push_back(i - 1);
              a[i - 1] = a[i + 1] = 0;
              work(1, i);
              work(i, n);
              break;
            }
            i = j + 1;
          }
        if (i > n) {
          puts("NO");
          continue;
        }
      }
    }
    puts("YES");
    cout << ans.size() << endl;
    for (auto x : ans) printf("%d ", x);
    puts("");
  }
}

```