#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 100;
int a[maxn];
int n;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    int jud = 0;
    for (int i = 1; i <= n; ++i)
      if (a[i] == 1) ++jud;
    if (jud & 1) {
      printf("NO\n");
      continue;
    }
    vector<int> res;
    for (int i = 1; i + 2 <= n;) {
      if (a[i] == 1) {
        int cnt = 1 + a[i + 1] + a[i + 2];
        if (cnt & 1) {
          if (cnt != 3) {
            res.push_back(i);
            a[i] = a[i + 1] = a[i + 2] = 1;
          }
          i = i + 2;
        } else {
          res.push_back(i);
          a[i] = a[i + 1] = a[i + 2] = 0;
          i = i + 3;
        }
      } else
        ++i;
    }
    for (int i = 1; i + 2 <= n; ++i) {
      int cnt = a[i] + a[i + 1] + a[i + 2];
      if (cnt == 2) {
        res.push_back(i);
        a[i] = a[i + 1] = a[i + 2] = 0;
      }
    }
    for (int i = n; i - 2 >= 1; --i) {
      int cnt = a[i] + a[i - 1] + a[i - 2];
      if (cnt == 2) {
        res.push_back(i - 2);
        a[i] = a[i - 1] = a[i - 2] = 0;
      }
    }
    bool f = true;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 1) {
        f = false;
        break;
      }
    }
    if (!f) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    printf("%d\n", (int)res.size());
    for (int num : res) {
      printf("%d ", num);
    }
    puts("");
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 maxn = 2e5 + 100;
int a[maxn];
int n;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    int jud = 0;
    for (int i = 1; i <= n; ++i)
      if (a[i] == 1) ++jud;
    if (jud & 1) {
      printf("NO\n");
      continue;
    }
    vector<int> res;
    for (int i = 1; i + 2 <= n;) {
      if (a[i] == 1) {
        int cnt = 1 + a[i + 1] + a[i + 2];
        if (cnt & 1) {
          if (cnt != 3) {
            res.push_back(i);
            a[i] = a[i + 1] = a[i + 2] = 1;
          }
          i = i + 2;
        } else {
          res.push_back(i);
          a[i] = a[i + 1] = a[i + 2] = 0;
          i = i + 3;
        }
      } else
        ++i;
    }
    for (int i = 1; i + 2 <= n; ++i) {
      int cnt = a[i] + a[i + 1] + a[i + 2];
      if (cnt == 2) {
        res.push_back(i);
        a[i] = a[i + 1] = a[i + 2] = 0;
      }
    }
    for (int i = n; i - 2 >= 1; --i) {
      int cnt = a[i] + a[i - 1] + a[i - 2];
      if (cnt == 2) {
        res.push_back(i - 2);
        a[i] = a[i - 1] = a[i - 2] = 0;
      }
    }
    bool f = true;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 1) {
        f = false;
        break;
      }
    }
    if (!f) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    printf("%d\n", (int)res.size());
    for (int num : res) {
      printf("%d ", num);
    }
    puts("");
  }
}

```