#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
int N;
int A[200011];
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
  int cnt = count(A, A + N, 1);
  bool yes = true;
  if (cnt % 2) yes = false;
  if (yes) {
    if (N % 2 == 0 && A[0] == 1 && A[N - 1] == 1) {
      bool p = true;
      for (int i = 1; i + 1 < N; i += 2) {
        if (A[i] != A[i + 1]) p = false;
      }
      if (p) yes = false;
    }
  }
  if (yes) {
    vector<int> ans;
    for (int i = 0; i < N - 2;) {
      int m = A[i] + A[i + 1] + A[i + 2];
      if (m == 0) {
        i++;
      } else if (m == 1) {
        if (A[i] == 1) {
          ans.push_back(i + 1);
          A[i] = A[i + 1] = A[i + 2] = 1;
        } else {
          i++;
        }
      } else if (m == 2) {
        ans.push_back(i + 1);
        A[i] = A[i + 1] = A[i + 2] = 0;
        i = max(0, i - 2);
      } else if (m == 3) {
        if (i && A[i - 1] == 0)
          i--;
        else
          i += 2;
      } else {
        assert(false);
      }
    }
    assert(count(A, A + N, 1) == 0);
    puts("YES");
    printf("%d\n", (int)ans.size());
    if (ans.size()) {
      do {
        const auto end_rp = (ans.end());
        auto it_rp = (ans.begin());
        for (bool sp_rp = 0; it_rp != end_rp; ++it_rp) {
          if (sp_rp)
            putchar(' ');
          else
            sp_rp = true;
          printf("%d", *it_rp);
        }
        putchar('\n');
      } while (0);
    }
  } else {
    puts("NO");
  }
}
int main() {
  int TC = 1;
  scanf("%d", &TC);
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
int N;
int A[200011];
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
  int cnt = count(A, A + N, 1);
  bool yes = true;
  if (cnt % 2) yes = false;
  if (yes) {
    if (N % 2 == 0 && A[0] == 1 && A[N - 1] == 1) {
      bool p = true;
      for (int i = 1; i + 1 < N; i += 2) {
        if (A[i] != A[i + 1]) p = false;
      }
      if (p) yes = false;
    }
  }
  if (yes) {
    vector<int> ans;
    for (int i = 0; i < N - 2;) {
      int m = A[i] + A[i + 1] + A[i + 2];
      if (m == 0) {
        i++;
      } else if (m == 1) {
        if (A[i] == 1) {
          ans.push_back(i + 1);
          A[i] = A[i + 1] = A[i + 2] = 1;
        } else {
          i++;
        }
      } else if (m == 2) {
        ans.push_back(i + 1);
        A[i] = A[i + 1] = A[i + 2] = 0;
        i = max(0, i - 2);
      } else if (m == 3) {
        if (i && A[i - 1] == 0)
          i--;
        else
          i += 2;
      } else {
        assert(false);
      }
    }
    assert(count(A, A + N, 1) == 0);
    puts("YES");
    printf("%d\n", (int)ans.size());
    if (ans.size()) {
      do {
        const auto end_rp = (ans.end());
        auto it_rp = (ans.begin());
        for (bool sp_rp = 0; it_rp != end_rp; ++it_rp) {
          if (sp_rp)
            putchar(' ');
          else
            sp_rp = true;
          printf("%d", *it_rp);
        }
        putchar('\n');
      } while (0);
    }
  } else {
    puts("NO");
  }
}
int main() {
  int TC = 1;
  scanf("%d", &TC);
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

```