#include <bits/stdc++.h>
using namespace std;
const long long MOD = (long long)1e9 + 7;
const int SIZE = 1000002;
const long long INF = 1LL << 60;
const double eps = 1e-6;
const double PI = 3.1415926535897932;
int a[200009];
int rs[200009];
int n;
vector<int> ans;
int main() {
  int ___T, case_n = 1;
  scanf("%d ", &___T);
  while (___T-- > 0) {
    scanf("%d", &(n));
    int oddz = -1;
    for (int i = 0; i < (n); ++i) {
      scanf("%d", &(a[i]));
      rs[i + 1] = rs[i] ^ a[i];
      if ((i + 1) % 2 && rs[i + 1] == 0) {
        oddz = i + 1;
      }
    }
    if (oddz == -1 || rs[n]) {
      puts("NO");
      continue;
    }
    if (oddz != -1) {
      ans.clear();
      for (int i = oddz + 1; i + 2 <= n; i += 2) {
        ans.push_back(i);
      }
      for (int i = oddz - 2; i >= 1; i -= 2) {
        ans.push_back(i);
      }
    }
    for (int i = 1; i + 2 <= n; i += 2) {
      ans.push_back(i);
    }
    puts("YES");
    printf("%d\n", ((int)(ans).size()));
    for (int i = 0; i < (((int)(ans).size())); ++i) {
      if (i) printf(" ");
      printf("%d", ans[i]);
    }
    puts("");
    case_n++;
  }
}

### 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>
using namespace std;
const long long MOD = (long long)1e9 + 7;
const int SIZE = 1000002;
const long long INF = 1LL << 60;
const double eps = 1e-6;
const double PI = 3.1415926535897932;
int a[200009];
int rs[200009];
int n;
vector<int> ans;
int main() {
  int ___T, case_n = 1;
  scanf("%d ", &___T);
  while (___T-- > 0) {
    scanf("%d", &(n));
    int oddz = -1;
    for (int i = 0; i < (n); ++i) {
      scanf("%d", &(a[i]));
      rs[i + 1] = rs[i] ^ a[i];
      if ((i + 1) % 2 && rs[i + 1] == 0) {
        oddz = i + 1;
      }
    }
    if (oddz == -1 || rs[n]) {
      puts("NO");
      continue;
    }
    if (oddz != -1) {
      ans.clear();
      for (int i = oddz + 1; i + 2 <= n; i += 2) {
        ans.push_back(i);
      }
      for (int i = oddz - 2; i >= 1; i -= 2) {
        ans.push_back(i);
      }
    }
    for (int i = 1; i + 2 <= n; i += 2) {
      ans.push_back(i);
    }
    puts("YES");
    printf("%d\n", ((int)(ans).size()));
    for (int i = 0; i < (((int)(ans).size())); ++i) {
      if (i) printf(" ");
      printf("%d", ans[i]);
    }
    puts("");
    case_n++;
  }
}

```