#include <bits/stdc++.h>
using namespace std;
const int N = 222222;
int n, a[N];
vector<int> v, s;
void solver(int x, int y) {
  while (y - x >= 3) {
    s.push_back(y - 2);
    s.push_back(y - 1);
    y -= 2;
  }
  s.push_back(x);
}
void solvel(int x, int y) {
  while (y - x >= 3) {
    s.push_back(x);
    s.push_back(x - 1);
    x += 2;
  }
  s.push_back(y - 2);
}
void solvem(int x, int y) {
  int i;
  for (i = x; i < y; i += 2) s.push_back(i);
  for (i = y - 4; i >= x; i -= 2) s.push_back(i);
}
int main() {
  int T, i, x;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    x = 0;
    s.clear();
    v.clear();
    for (i = 1; i <= n; i++) {
      scanf("%d", a + i);
      x ^= a[i];
      if (a[i]) v.push_back(i);
    }
    if (x) {
      printf("NO\n");
      continue;
    }
    if (!a[n]) {
      for (i = v.size() - 1; i >= 0; i -= 2) solver(v[i - 1], v[i]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    if (!a[1]) {
      for (i = 0; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    x = -1;
    for (i = 1; i < v.size() - 1; i += 2)
      if (v[i] < v[i + 1] - 1) x = i;
    if (x >= 0) {
      for (i = x; i >= 0; i -= 2) solver(v[i - 1], v[i]);
      for (i = x + 1; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    x = -1;
    for (i = 0; i < v.size() - 1; i++)
      if (v[i] % 2 == v[i + 1] % 2) x = i;
    if (x >= 0) {
      solvem(v[x], v[x + 1]);
      for (i = x - 1; i >= 0; i -= 2) solver(v[i - 1], v[i]);
      for (i = x + 2; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 N = 222222;
int n, a[N];
vector<int> v, s;
void solver(int x, int y) {
  while (y - x >= 3) {
    s.push_back(y - 2);
    s.push_back(y - 1);
    y -= 2;
  }
  s.push_back(x);
}
void solvel(int x, int y) {
  while (y - x >= 3) {
    s.push_back(x);
    s.push_back(x - 1);
    x += 2;
  }
  s.push_back(y - 2);
}
void solvem(int x, int y) {
  int i;
  for (i = x; i < y; i += 2) s.push_back(i);
  for (i = y - 4; i >= x; i -= 2) s.push_back(i);
}
int main() {
  int T, i, x;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    x = 0;
    s.clear();
    v.clear();
    for (i = 1; i <= n; i++) {
      scanf("%d", a + i);
      x ^= a[i];
      if (a[i]) v.push_back(i);
    }
    if (x) {
      printf("NO\n");
      continue;
    }
    if (!a[n]) {
      for (i = v.size() - 1; i >= 0; i -= 2) solver(v[i - 1], v[i]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    if (!a[1]) {
      for (i = 0; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    x = -1;
    for (i = 1; i < v.size() - 1; i += 2)
      if (v[i] < v[i + 1] - 1) x = i;
    if (x >= 0) {
      for (i = x; i >= 0; i -= 2) solver(v[i - 1], v[i]);
      for (i = x + 1; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    x = -1;
    for (i = 0; i < v.size() - 1; i++)
      if (v[i] % 2 == v[i + 1] % 2) x = i;
    if (x >= 0) {
      solvem(v[x], v[x + 1]);
      for (i = x - 1; i >= 0; i -= 2) solver(v[i - 1], v[i]);
      for (i = x + 2; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
      printf("YES\n");
      printf("%d\n", s.size());
      for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
      printf("\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}

```