#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
template <typename T>
inline void read(T &WOW) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  WOW = flag * x;
}
int n, a[MAXN];
vector<int> ans;
void Construct(int l, int r) {
  for (int i = l; i <= r - 2; i += 2) {
    ans.push_back(i);
  }
  for (int i = r - 4; i >= l; i -= 2) {
    ans.push_back(i);
  }
}
void solve() {
  read(n);
  int xorsum = 0;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    xorsum ^= a[i];
  }
  if (xorsum) {
    puts("NO");
    return;
  }
  ans.clear();
  if (n & 1) {
    Construct(1, n);
  } else {
    xorsum = 0;
    bool flag = 0;
    for (int i = 1; i <= n; ++i) {
      xorsum ^= a[i];
      if (!xorsum && (i & 1)) {
        Construct(1, i);
        Construct(i + 1, n);
        flag = 1;
        break;
      }
    }
    if (!flag) {
      puts("NO");
      return;
    }
  }
  puts("YES");
  printf("%d\n", ans.size());
  for (auto x : ans) {
    printf("%d ", x);
  }
  printf("\n");
}
int main() {
  int T;
  read(T);
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 200005;
template <typename T>
inline void read(T &WOW) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  WOW = flag * x;
}
int n, a[MAXN];
vector<int> ans;
void Construct(int l, int r) {
  for (int i = l; i <= r - 2; i += 2) {
    ans.push_back(i);
  }
  for (int i = r - 4; i >= l; i -= 2) {
    ans.push_back(i);
  }
}
void solve() {
  read(n);
  int xorsum = 0;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    xorsum ^= a[i];
  }
  if (xorsum) {
    puts("NO");
    return;
  }
  ans.clear();
  if (n & 1) {
    Construct(1, n);
  } else {
    xorsum = 0;
    bool flag = 0;
    for (int i = 1; i <= n; ++i) {
      xorsum ^= a[i];
      if (!xorsum && (i & 1)) {
        Construct(1, i);
        Construct(i + 1, n);
        flag = 1;
        break;
      }
    }
    if (!flag) {
      puts("NO");
      return;
    }
  }
  puts("YES");
  printf("%d\n", ans.size());
  for (auto x : ans) {
    printf("%d ", x);
  }
  printf("\n");
}
int main() {
  int T;
  read(T);
  while (T--) {
    solve();
  }
  return 0;
}

```