#include <bits/stdc++.h>
using namespace std;
vector<long long> solveoddarray(vector<long long>& a, long long start,
                                long long end) {
  vector<long long> ans;
  for (long long i = start; i <= end - 2; i += 2) {
    ans.push_back(i + 1);
  }
  for (long long i = end - 4; i >= start; i -= 2) {
    ans.push_back(i + 1);
  }
  return ans;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    scanf("%lld", &n);
    vector<long long> a(n);
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      sum += a[i];
    }
    bool possible = true;
    vector<long long> ans;
    if (sum % 2 == 0) {
      if (n % 2 == 1) {
        ans = solveoddarray(a, 0, n - 1);
      } else {
        if (a[0] == 0) {
          ans = solveoddarray(a, 1, n - 1);
        } else {
          long long s = 0;
          possible = false;
          for (long long i = 0; i < n; i++) {
            s += a[i];
            vector<long long> t1;
            if (s % 2 == 0 && i % 2 == 0) {
              ans = solveoddarray(a, 0, i);
              t1 = solveoddarray(a, i + 1, n - 1);
              for (long long j = 0; j < (long long)t1.size(); j++) {
                ans.push_back(t1[j]);
              }
              possible = true;
              break;
            }
          }
        }
      }
    } else {
      possible = false;
    }
    if (!possible) {
      printf("NO\n");
    } else {
      printf("YES\n");
      printf("%lld\n", (long long)ans.size());
      if ((long long)ans.size() > 0) {
        for (long long i = 0; i < (long long)ans.size(); i++) {
          printf("%lld ", ans[i]);
        }
        printf("\n");
      }
    }
  }
}

### Prompt

Your task is to create 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;
vector<long long> solveoddarray(vector<long long>& a, long long start,
                                long long end) {
  vector<long long> ans;
  for (long long i = start; i <= end - 2; i += 2) {
    ans.push_back(i + 1);
  }
  for (long long i = end - 4; i >= start; i -= 2) {
    ans.push_back(i + 1);
  }
  return ans;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    scanf("%lld", &n);
    vector<long long> a(n);
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      sum += a[i];
    }
    bool possible = true;
    vector<long long> ans;
    if (sum % 2 == 0) {
      if (n % 2 == 1) {
        ans = solveoddarray(a, 0, n - 1);
      } else {
        if (a[0] == 0) {
          ans = solveoddarray(a, 1, n - 1);
        } else {
          long long s = 0;
          possible = false;
          for (long long i = 0; i < n; i++) {
            s += a[i];
            vector<long long> t1;
            if (s % 2 == 0 && i % 2 == 0) {
              ans = solveoddarray(a, 0, i);
              t1 = solveoddarray(a, i + 1, n - 1);
              for (long long j = 0; j < (long long)t1.size(); j++) {
                ans.push_back(t1[j]);
              }
              possible = true;
              break;
            }
          }
        }
      }
    } else {
      possible = false;
    }
    if (!possible) {
      printf("NO\n");
    } else {
      printf("YES\n");
      printf("%lld\n", (long long)ans.size());
      if ((long long)ans.size() > 0) {
        for (long long i = 0; i < (long long)ans.size(); i++) {
          printf("%lld ", ans[i]);
        }
        printf("\n");
      }
    }
  }
}

```