#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
int a, b, c, d, n, m, k, kolt;
int mas[200002];
int main() {
  scanf("%d", &kolt);
  for (int _n((kolt)), hod(1); hod <= _n; hod++) {
    scanf("%d", &n);
    c = 0;
    b = 0;
    for (int _n((n)-1), i(0); i <= _n; i++) {
      scanf("%d", &mas[i]);
      c ^= mas[i];
      b += mas[i];
    }
    if (c || b == n) {
      printf("NO\n");
      continue;
    }
    vector<int> res;
    bool bad = 0;
    for (int _n((n)-1), i(0); i <= _n; i++) {
      if (!mas[i]) continue;
      if (i && mas[i - 1] == 1) continue;
      int a = i;
      while (a < n && mas[a] == 1) ++a;
      while ((a - i) % 2 == 1) {
        b = a;
        while (mas[b] == 0) ++b;
        for (int j = b - 2; j >= a; j -= 2) {
          res.push_back(j);
          mas[j] = mas[j + 1] = mas[j + 2] = 1;
          b -= 2;
        }
        if (b == a + 1) {
          res.push_back(a - 1);
          mas[a - 1] = mas[a] = mas[a + 1] = 0;
          --a;
          break;
        } else {
          while (a < n && mas[a] == 1) ++a;
        }
      }
      if (a <= i) continue;
      if (i - 1 >= 0 && mas[i - 1] == 0) {
        for (int j = i; j < a; j += 2) {
          res.push_back(j - 1);
          mas[j - 1] = mas[j] = mas[j + 1] = 0;
        }
      } else if (a < n) {
        for (int j = a - 2; j >= i; j -= 2) {
          res.push_back(j);
          mas[j] = mas[j + 1] = mas[j + 2] = 0;
        }
      } else {
        bad = 1;
        break;
      }
    }
    if (bad) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    printf("%d\n", (int)((res).size()));
    if (res.empty()) continue;
    for (int _n(((int)((res).size())) - 1), i(0); i <= _n; i++) {
      if (i) printf(" ");
      printf("%d", res[i] + 1);
    }
    printf("\n");
  }
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
int a, b, c, d, n, m, k, kolt;
int mas[200002];
int main() {
  scanf("%d", &kolt);
  for (int _n((kolt)), hod(1); hod <= _n; hod++) {
    scanf("%d", &n);
    c = 0;
    b = 0;
    for (int _n((n)-1), i(0); i <= _n; i++) {
      scanf("%d", &mas[i]);
      c ^= mas[i];
      b += mas[i];
    }
    if (c || b == n) {
      printf("NO\n");
      continue;
    }
    vector<int> res;
    bool bad = 0;
    for (int _n((n)-1), i(0); i <= _n; i++) {
      if (!mas[i]) continue;
      if (i && mas[i - 1] == 1) continue;
      int a = i;
      while (a < n && mas[a] == 1) ++a;
      while ((a - i) % 2 == 1) {
        b = a;
        while (mas[b] == 0) ++b;
        for (int j = b - 2; j >= a; j -= 2) {
          res.push_back(j);
          mas[j] = mas[j + 1] = mas[j + 2] = 1;
          b -= 2;
        }
        if (b == a + 1) {
          res.push_back(a - 1);
          mas[a - 1] = mas[a] = mas[a + 1] = 0;
          --a;
          break;
        } else {
          while (a < n && mas[a] == 1) ++a;
        }
      }
      if (a <= i) continue;
      if (i - 1 >= 0 && mas[i - 1] == 0) {
        for (int j = i; j < a; j += 2) {
          res.push_back(j - 1);
          mas[j - 1] = mas[j] = mas[j + 1] = 0;
        }
      } else if (a < n) {
        for (int j = a - 2; j >= i; j -= 2) {
          res.push_back(j);
          mas[j] = mas[j + 1] = mas[j + 2] = 0;
        }
      } else {
        bad = 1;
        break;
      }
    }
    if (bad) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    printf("%d\n", (int)((res).size()));
    if (res.empty()) continue;
    for (int _n(((int)((res).size())) - 1), i(0); i <= _n; i++) {
      if (i) printf(" ");
      printf("%d", res[i] + 1);
    }
    printf("\n");
  }
}

```