#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
const int inf = 2147483647;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10ll + ch - '0', ch = getchar();
  return x * f;
}
int n, a[N];
vector<int> ans;
void solve() {
  n = read();
  a[0] = 0;
  for (int i = 1; i <= n; i++) a[i] = read() ^ a[i - 1];
  int p = -1;
  for (int i = 1; i <= n; i += 2)
    if (!a[i]) {
      p = i;
      break;
    }
  if (p == -1 || a[n]) {
    puts("NO");
    return;
  }
  puts("YES");
  ans.clear();
  for (int i = p + 1; i + 2 <= n; i += 2) ans.push_back(i);
  for (int i = p - 2; i > 0; i -= 2) ans.push_back(i);
  for (int i = 1; i + 2 <= n; i += 2) ans.push_back(i);
  printf("%d\n", ans.size());
  for (int x : ans) printf("%d ", x);
  puts("");
}
int main() {
  int T = read();
  while (T--) solve();
}

### 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;
const int N = 200010;
const int inf = 2147483647;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10ll + ch - '0', ch = getchar();
  return x * f;
}
int n, a[N];
vector<int> ans;
void solve() {
  n = read();
  a[0] = 0;
  for (int i = 1; i <= n; i++) a[i] = read() ^ a[i - 1];
  int p = -1;
  for (int i = 1; i <= n; i += 2)
    if (!a[i]) {
      p = i;
      break;
    }
  if (p == -1 || a[n]) {
    puts("NO");
    return;
  }
  puts("YES");
  ans.clear();
  for (int i = p + 1; i + 2 <= n; i += 2) ans.push_back(i);
  for (int i = p - 2; i > 0; i -= 2) ans.push_back(i);
  for (int i = 1; i + 2 <= n; i += 2) ans.push_back(i);
  printf("%d\n", ans.size());
  for (int x : ans) printf("%d ", x);
  puts("");
}
int main() {
  int T = read();
  while (T--) solve();
}

```