#include <bits/stdc++.h>
using namespace std;
template <typename... Args>
void trace(Args... args) {}
long long t = 1, n, ans, a[1001000];
int main() {
  scanf("%lld", &t);
st:
  ans = 0;
  scanf("%lld", &n);
  long long xo = 0, f[] = {0, 0};
  vector<int> ones;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    f[a[i]]++;
    xo ^= a[i];
    if (a[i]) ones.push_back(i);
  }
  if (xo == 1 || f[0] == 0)
    puts("NO");
  else {
    vector<int> op;
    int l = 0, r = 0;
    while (1) {
      while (r < n) {
        if (a[l] == 0) {
          l++;
          r = l;
        } else if (a[r] == 1) {
          r++;
        } else
          break;
      }
      if (l == n)
        break;
      else if (l == 0 && r == n)
        break;
      else if ((r - l) % 2 == 0) {
        if (l > 0) {
          while (l < r) {
            op.push_back(l);
            assert(a[l - 1] == 0 && a[l] == 1 && a[l + 1] == 1);
            a[l - 1] = a[l] = a[l + 1] = 0;
            l += 2;
          }
        } else {
          for (int j = r; j > l; j -= 2) {
            op.push_back(j - 1);
            assert(a[j - 2] == 1 && a[j - 1] == 1 && a[j] == 0);
            a[j - 2] = a[j - 1] = a[j] = 0;
          }
        }
      } else {
        if (r + 1 >= n) break;
        long long tmp = a[r - 1] ^ a[r] ^ a[r + 1];
        a[r - 1] = a[r] = a[r + 1] = tmp;
        op.push_back(r);
        if (tmp == 0) r--;
      }
    }
    int can = 1;
    for (int i = 0; i < n; i++) can &= (a[i] == 0);
    if (!can)
      puts("NO");
    else {
      printf("YES\n%d\n", (int)op.size());
      for (auto x : op) printf("%d ", x);
      puts("");
    }
  }
  if (--t) goto st;
  return 0;
}

### Prompt

Generate 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;
template <typename... Args>
void trace(Args... args) {}
long long t = 1, n, ans, a[1001000];
int main() {
  scanf("%lld", &t);
st:
  ans = 0;
  scanf("%lld", &n);
  long long xo = 0, f[] = {0, 0};
  vector<int> ones;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    f[a[i]]++;
    xo ^= a[i];
    if (a[i]) ones.push_back(i);
  }
  if (xo == 1 || f[0] == 0)
    puts("NO");
  else {
    vector<int> op;
    int l = 0, r = 0;
    while (1) {
      while (r < n) {
        if (a[l] == 0) {
          l++;
          r = l;
        } else if (a[r] == 1) {
          r++;
        } else
          break;
      }
      if (l == n)
        break;
      else if (l == 0 && r == n)
        break;
      else if ((r - l) % 2 == 0) {
        if (l > 0) {
          while (l < r) {
            op.push_back(l);
            assert(a[l - 1] == 0 && a[l] == 1 && a[l + 1] == 1);
            a[l - 1] = a[l] = a[l + 1] = 0;
            l += 2;
          }
        } else {
          for (int j = r; j > l; j -= 2) {
            op.push_back(j - 1);
            assert(a[j - 2] == 1 && a[j - 1] == 1 && a[j] == 0);
            a[j - 2] = a[j - 1] = a[j] = 0;
          }
        }
      } else {
        if (r + 1 >= n) break;
        long long tmp = a[r - 1] ^ a[r] ^ a[r + 1];
        a[r - 1] = a[r] = a[r + 1] = tmp;
        op.push_back(r);
        if (tmp == 0) r--;
      }
    }
    int can = 1;
    for (int i = 0; i < n; i++) can &= (a[i] == 0);
    if (!can)
      puts("NO");
    else {
      printf("YES\n%d\n", (int)op.size());
      for (auto x : op) printf("%d ", x);
      puts("");
    }
  }
  if (--t) goto st;
  return 0;
}

```