#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    w |= ch == '-';
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return w ? -x : x;
}
inline long long readll() {
  long long x = 0, w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    w |= ch == '-';
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return w ? -x : x;
}
int Tc;
int n, a[200005], flag;
int ans[200005], cnt;
void work(int x) {
  ans[++cnt] = x;
  int tmp = a[x] ^ a[x + 1] ^ a[x + 2];
  a[x] = a[x + 1] = a[x + 2] = tmp;
}
void solve() {
  cnt = 0;
  scanf("%d", &n);
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    sum += a[i];
  }
  if (sum & 1) {
    printf("NO\n");
    return;
  }
  sum = 0, flag = 0;
  for (int i = 1; i <= n; i++) {
    sum ^= a[i];
    if (sum == 0 && a[i] == 0) flag = i;
  }
  if (!flag) {
    int last = 1;
    for (int i = 1; i <= n; i++) {
      if (a[i] && ((i - last) & 1)) flag = i - 1;
      if (a[i]) last = i + 1;
    }
    if (!flag) {
      printf("NO\n");
      return;
    }
    for (; !a[flag - 1]; flag -= 2) work(flag - 1);
    work(flag - 1);
    flag--;
  }
  for (int i = flag - 1; i >= 1; i--) {
    if (!a[i]) continue;
    if (a[i - 1])
      work(i - 1);
    else {
      work(i - 2);
      if (a[i] == 1) work(i - 1);
    }
  }
  for (int i = flag + 1; i <= n; i++) {
    if (!a[i]) continue;
    if (a[i + 1])
      work(i - 1);
    else {
      work(i);
      if (a[i] == 1) work(i - 1);
    }
  }
  printf("YES\n%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  Tc = read();
  while (Tc--) solve();
  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;
inline int read() {
  int x = 0, w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    w |= ch == '-';
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return w ? -x : x;
}
inline long long readll() {
  long long x = 0, w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    w |= ch == '-';
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return w ? -x : x;
}
int Tc;
int n, a[200005], flag;
int ans[200005], cnt;
void work(int x) {
  ans[++cnt] = x;
  int tmp = a[x] ^ a[x + 1] ^ a[x + 2];
  a[x] = a[x + 1] = a[x + 2] = tmp;
}
void solve() {
  cnt = 0;
  scanf("%d", &n);
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    sum += a[i];
  }
  if (sum & 1) {
    printf("NO\n");
    return;
  }
  sum = 0, flag = 0;
  for (int i = 1; i <= n; i++) {
    sum ^= a[i];
    if (sum == 0 && a[i] == 0) flag = i;
  }
  if (!flag) {
    int last = 1;
    for (int i = 1; i <= n; i++) {
      if (a[i] && ((i - last) & 1)) flag = i - 1;
      if (a[i]) last = i + 1;
    }
    if (!flag) {
      printf("NO\n");
      return;
    }
    for (; !a[flag - 1]; flag -= 2) work(flag - 1);
    work(flag - 1);
    flag--;
  }
  for (int i = flag - 1; i >= 1; i--) {
    if (!a[i]) continue;
    if (a[i - 1])
      work(i - 1);
    else {
      work(i - 2);
      if (a[i] == 1) work(i - 1);
    }
  }
  for (int i = flag + 1; i <= n; i++) {
    if (!a[i]) continue;
    if (a[i + 1])
      work(i - 1);
    else {
      work(i);
      if (a[i] == 1) work(i - 1);
    }
  }
  printf("YES\n%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  Tc = read();
  while (Tc--) solve();
  return 0;
}

```