#include <bits/stdc++.h>
using namespace std;
int n;
int a[200200];
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int pref = 0;
  for (int i = 0; i < n; i++) {
    if (a[i])
      pref++;
    else
      break;
  }
  if (pref == n) {
    printf("NO\n");
    return;
  }
  vector<int> rj;
  if (pref % 2 == 1) {
    for (int i = pref - 1; i < n - 2; i += 1) {
      int xr = a[i] ^ a[i + 1] ^ a[i + 2];
      int sm = a[i] + a[i + 1] + a[i + 2];
      if (xr == 0) {
        rj.push_back(i + 1);
        a[i] = 0;
        a[i + 1] = 0;
        a[i + 2] = 0;
        pref -= 1;
        break;
      } else {
        if (sm == 1) {
          rj.push_back(i + 1);
          a[i + 1] = 1;
          a[i + 2] = 1;
          i += 1;
          pref += 2;
        } else if (sm == 3) {
          i += 1;
          pref += 2;
        }
      }
    }
  }
  if (pref >= n) {
    printf("NO\n");
    return;
  }
  if (pref % 2 == 1) {
    printf("NO\n");
    return;
  }
  for (int i = pref - 1; i >= 1; i -= 2) {
    rj.push_back(i);
  }
  for (int i = 0; i < pref; i++) a[i] = 0;
  for (int i = pref; i < n - 2; i++) {
    int xr = a[i] ^ a[i + 1] ^ a[i + 2];
    int sm = a[i] + a[i + 1] + a[i + 2];
    if (sm == 0) continue;
    if (a[i]) {
      if (xr == 1) {
        rj.push_back(i + 1);
        rj.push_back(i);
        a[i] = 0;
        i++;
        a[i + 1] = 1;
        a[i] = 0;
        i--;
      } else {
        rj.push_back(i + 1);
        a[i] = 0;
        a[i + 1] = 0;
        a[i + 2] = 0;
      }
    } else {
      if (xr == 0) {
        rj.push_back(i + 1);
        a[i] = 0;
        a[i + 1] = 0;
        a[i + 2] = 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (a[i]) {
      printf("NO\n");
      return;
    }
  }
  printf("YES\n%d\n", rj.size());
  for (int i : rj) {
    printf("%d ", i);
  }
  printf("\n");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int n;
int a[200200];
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int pref = 0;
  for (int i = 0; i < n; i++) {
    if (a[i])
      pref++;
    else
      break;
  }
  if (pref == n) {
    printf("NO\n");
    return;
  }
  vector<int> rj;
  if (pref % 2 == 1) {
    for (int i = pref - 1; i < n - 2; i += 1) {
      int xr = a[i] ^ a[i + 1] ^ a[i + 2];
      int sm = a[i] + a[i + 1] + a[i + 2];
      if (xr == 0) {
        rj.push_back(i + 1);
        a[i] = 0;
        a[i + 1] = 0;
        a[i + 2] = 0;
        pref -= 1;
        break;
      } else {
        if (sm == 1) {
          rj.push_back(i + 1);
          a[i + 1] = 1;
          a[i + 2] = 1;
          i += 1;
          pref += 2;
        } else if (sm == 3) {
          i += 1;
          pref += 2;
        }
      }
    }
  }
  if (pref >= n) {
    printf("NO\n");
    return;
  }
  if (pref % 2 == 1) {
    printf("NO\n");
    return;
  }
  for (int i = pref - 1; i >= 1; i -= 2) {
    rj.push_back(i);
  }
  for (int i = 0; i < pref; i++) a[i] = 0;
  for (int i = pref; i < n - 2; i++) {
    int xr = a[i] ^ a[i + 1] ^ a[i + 2];
    int sm = a[i] + a[i + 1] + a[i + 2];
    if (sm == 0) continue;
    if (a[i]) {
      if (xr == 1) {
        rj.push_back(i + 1);
        rj.push_back(i);
        a[i] = 0;
        i++;
        a[i + 1] = 1;
        a[i] = 0;
        i--;
      } else {
        rj.push_back(i + 1);
        a[i] = 0;
        a[i + 1] = 0;
        a[i + 2] = 0;
      }
    } else {
      if (xr == 0) {
        rj.push_back(i + 1);
        a[i] = 0;
        a[i + 1] = 0;
        a[i + 2] = 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (a[i]) {
      printf("NO\n");
      return;
    }
  }
  printf("YES\n%d\n", rj.size());
  for (int i : rj) {
    printf("%d ", i);
  }
  printf("\n");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}

```