#include <bits/stdc++.h>
int i, j, k, n, m, t, a[200500], res, s[200500], it;
void add(int i, int x) {
  s[++it] = i;
  a[i] = a[i + 1] = a[i + 2] = x;
}
bool chk(int i) {
  if (i + 2 > n) return 0;
  if ((a[i] + a[i + 1] + a[i + 2]) == 2) add(i, 0);
  if (!a[i]) return 1;
  if (a[i])
    if (!(a[i + 1] + a[i + 2])) add(i, 1);
  if (chk(i + 2)) {
    add(i, 0);
    return 1;
  }
  return 0;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    k = 0;
    it = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      k += a[i];
    }
    if (k & 1) {
      puts("NO");
      continue;
    }
    for (i = 1; i <= n - 2; i++) {
      if (!chk(i)) break;
    }
    for (i = 1; i <= n - 2; i++) {
      if (!chk(i)) {
        puts("NO");
        goto aaa;
      }
    }
    printf("YES\n%d\n", it);
    for (i = 1; i <= it; i++) {
      printf("%d ", s[i]);
    }
    puts("");
  aaa:;
  }
}

### Prompt

Please create a solution in Cpp 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>
int i, j, k, n, m, t, a[200500], res, s[200500], it;
void add(int i, int x) {
  s[++it] = i;
  a[i] = a[i + 1] = a[i + 2] = x;
}
bool chk(int i) {
  if (i + 2 > n) return 0;
  if ((a[i] + a[i + 1] + a[i + 2]) == 2) add(i, 0);
  if (!a[i]) return 1;
  if (a[i])
    if (!(a[i + 1] + a[i + 2])) add(i, 1);
  if (chk(i + 2)) {
    add(i, 0);
    return 1;
  }
  return 0;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    k = 0;
    it = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      k += a[i];
    }
    if (k & 1) {
      puts("NO");
      continue;
    }
    for (i = 1; i <= n - 2; i++) {
      if (!chk(i)) break;
    }
    for (i = 1; i <= n - 2; i++) {
      if (!chk(i)) {
        puts("NO");
        goto aaa;
      }
    }
    printf("YES\n%d\n", it);
    for (i = 1; i <= it; i++) {
      printf("%d ", s[i]);
    }
    puts("");
  aaa:;
  }
}

```