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].
#include <bits/stdc++.h>
using namespace std;
int a[200001];
int main() {
  int i, t;
  for (scanf("%d", &t); t--;) {
    int n;
    vector<int> ans;
    int cum = 0;
    for (scanf("%d", &n), i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      cum ^= a[i];
    }
    if (cum) {
      printf("NO\n");
      continue;
    } else if (cum == n) {
      printf("YES\n0\n");
      continue;
    }
    if (n & 1) {
      vector<int> ans;
      for (i = 1; i + 2 <= n; i += 2) ans.push_back(i);
      for (i = ans.back() - 2; i > 0; i -= 2) ans.push_back(i);
      printf("YES\n%d\n", ans.size());
      for (auto x : ans) printf("%d ", x);
      puts("");
    } else {
      cum = 0;
      int m = -1;
      int st = 1;
      int len = 1;
      while (a[st] == 0) ++st;
      for (i = st; i <= n; ++i, ++len) {
        cum ^= a[i];
        if (cum == 0 && (len & 1)) {
          m = i;
          break;
        }
      }
      if (m < 0)
        printf("NO\n");
      else {
        for (i = st; i + 2 <= m; i += 2) ans.push_back(i);
        for (i = ans.back() - 2; i > 0; i -= 2) ans.push_back(i);
        for (i = m + 1; i + 2 <= n; i += 2) ans.push_back(i);
        for (i = ans.back() - 2; i > m; i -= 2) ans.push_back(i);
        printf("YES\n%d\n", ans.size());
        for (auto x : ans) printf("%d ", x);
        puts("");
      }
    }
  }
  return 0;
}
