#include <bits/stdc++.h>
using namespace std;
const int inf = 1000000001;
const int MOD = 1000000007;
const int N = 200009;
int a[N + 1];
void doit(int pos, vector<int>& ans) {
  ans.push_back(pos);
  int t = a[pos] ^ a[pos + 1] ^ a[pos + 2];
  a[pos] = a[pos + 1] = a[pos + 2] = t;
}
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> ans;
  int tot = 0;
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    tot += a[i];
  }
  if (tot & 1) {
    cout << "NO" << endl;
    return;
  }
  int pr = 0;
  for (int i = (1); i <= (n); ++i) {
    if (!a[i]) {
      if (pr & 1) {
        int j = i + 1;
        while (j < n && !a[j]) j++;
        if (j == n + 1) {
          cout << "NO" << endl;
          return;
        }
        if ((j - i) % 2 == 0) {
          for (int f = j - 2; f >= i; f -= 2) doit(f, ans);
          pr += (j - i + 1);
          i = j;
          continue;
        }
        for (int f = j - 2; f > i; f -= 2) doit(f, ans);
        doit(i - 1, ans);
        for (int f = i - 3; f >= 1; f -= 2) {
          if (!a[f]) break;
          doit(f, ans);
        }
        for (int f = i + 1; f < j; f += 2) doit(f, ans);
        pr = 0;
      } else {
        for (int f = i - 2; f >= 1; f -= 2) {
          if (!a[f]) break;
          doit(f, ans);
        }
        pr = 0;
      }
    } else {
      pr++;
    }
  }
  if (pr & 1 || pr == n) {
    cout << "NO" << endl;
    return;
  }
  for (int f = n - pr; f < n; f += 2) doit(f, ans);
  cout << "YES" << endl;
  cout << (int)((ans).size()) << endl;
  for (int i = (0); i <= ((int)((ans).size()) - 1); ++i) {
    printf("%d ", ans[i]);
  }
  cout << endl;
}
int main(int argc, const char* argv[]) {
  int tests;
  scanf("%d\n", &tests);
  for (int i = (0); i <= (tests - 1); ++i) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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>
using namespace std;
const int inf = 1000000001;
const int MOD = 1000000007;
const int N = 200009;
int a[N + 1];
void doit(int pos, vector<int>& ans) {
  ans.push_back(pos);
  int t = a[pos] ^ a[pos + 1] ^ a[pos + 2];
  a[pos] = a[pos + 1] = a[pos + 2] = t;
}
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> ans;
  int tot = 0;
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    tot += a[i];
  }
  if (tot & 1) {
    cout << "NO" << endl;
    return;
  }
  int pr = 0;
  for (int i = (1); i <= (n); ++i) {
    if (!a[i]) {
      if (pr & 1) {
        int j = i + 1;
        while (j < n && !a[j]) j++;
        if (j == n + 1) {
          cout << "NO" << endl;
          return;
        }
        if ((j - i) % 2 == 0) {
          for (int f = j - 2; f >= i; f -= 2) doit(f, ans);
          pr += (j - i + 1);
          i = j;
          continue;
        }
        for (int f = j - 2; f > i; f -= 2) doit(f, ans);
        doit(i - 1, ans);
        for (int f = i - 3; f >= 1; f -= 2) {
          if (!a[f]) break;
          doit(f, ans);
        }
        for (int f = i + 1; f < j; f += 2) doit(f, ans);
        pr = 0;
      } else {
        for (int f = i - 2; f >= 1; f -= 2) {
          if (!a[f]) break;
          doit(f, ans);
        }
        pr = 0;
      }
    } else {
      pr++;
    }
  }
  if (pr & 1 || pr == n) {
    cout << "NO" << endl;
    return;
  }
  for (int f = n - pr; f < n; f += 2) doit(f, ans);
  cout << "YES" << endl;
  cout << (int)((ans).size()) << endl;
  for (int i = (0); i <= ((int)((ans).size()) - 1); ++i) {
    printf("%d ", ans[i]);
  }
  cout << endl;
}
int main(int argc, const char* argv[]) {
  int tests;
  scanf("%d\n", &tests);
  for (int i = (0); i <= (tests - 1); ++i) {
    solve();
  }
  return 0;
}

```