#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 2e5 + 5;
int t, n, a[N], c[N], tot, flag;
void solve() {
  n = read();
  tot = 0;
  flag = 1;
  for (int i = 1; i <= n; i++) a[i] = read();
  int i;
  if (a[1] == 0) {
    i = 2;
  } else {
    int j = 1;
    while ((a[j] == 1 && (a[j] + a[j + 1] + a[j + 2]) % 2 == 1) ||
           (a[j] == 0 && (a[j] + a[j + 1] + a[j + 2]) % 2 == 0)) {
      j += 2;
      if (j == n - 1) {
        flag = 0;
        break;
      }
      if (j == n) {
        flag = 0;
        break;
      }
    }
    if (flag == 0) {
      cout << "NO" << endl;
      return;
    } else {
      for (int k = j; k >= 1; k -= 2) c[++tot] = k;
      for (int k = 1; k <= j; k += 2)
        if (a[k] == 0) c[++tot] = k;
    }
    i = j + 3;
  }
  while (i <= n) {
    if (a[i] == 1) {
      if (i == n - 1) {
        if (a[n] == 1) {
          c[++tot] = i - 1;
          break;
        } else {
          flag = 0;
          break;
        }
      }
      if (i == n) {
        flag = 0;
        break;
      }
      if (a[i + 1] + a[i + 2] == 1) {
        c[++tot] = i;
        i += 3;
      } else if (a[i + 1] == 1) {
        c[++tot] = i - 1;
        i += 2;
      } else {
        c[++tot] = i;
        c[++tot] = i - 1;
        a[i + 2] = 1;
        i += 2;
      }
    } else {
      i++;
    }
  }
  if (flag == 0) {
    cout << "NO" << endl;
    return;
  } else {
    cout << "YES" << endl;
    cout << tot << endl;
    if (tot > 0) {
      for (int i = 1; i <= tot; i++) cout << c[i] << " ";
      cout << endl;
    }
  }
}
int main() {
  t = read();
  while (t--) {
    solve();
  }
  return 0;
}

### 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>
using namespace std;
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 2e5 + 5;
int t, n, a[N], c[N], tot, flag;
void solve() {
  n = read();
  tot = 0;
  flag = 1;
  for (int i = 1; i <= n; i++) a[i] = read();
  int i;
  if (a[1] == 0) {
    i = 2;
  } else {
    int j = 1;
    while ((a[j] == 1 && (a[j] + a[j + 1] + a[j + 2]) % 2 == 1) ||
           (a[j] == 0 && (a[j] + a[j + 1] + a[j + 2]) % 2 == 0)) {
      j += 2;
      if (j == n - 1) {
        flag = 0;
        break;
      }
      if (j == n) {
        flag = 0;
        break;
      }
    }
    if (flag == 0) {
      cout << "NO" << endl;
      return;
    } else {
      for (int k = j; k >= 1; k -= 2) c[++tot] = k;
      for (int k = 1; k <= j; k += 2)
        if (a[k] == 0) c[++tot] = k;
    }
    i = j + 3;
  }
  while (i <= n) {
    if (a[i] == 1) {
      if (i == n - 1) {
        if (a[n] == 1) {
          c[++tot] = i - 1;
          break;
        } else {
          flag = 0;
          break;
        }
      }
      if (i == n) {
        flag = 0;
        break;
      }
      if (a[i + 1] + a[i + 2] == 1) {
        c[++tot] = i;
        i += 3;
      } else if (a[i + 1] == 1) {
        c[++tot] = i - 1;
        i += 2;
      } else {
        c[++tot] = i;
        c[++tot] = i - 1;
        a[i + 2] = 1;
        i += 2;
      }
    } else {
      i++;
    }
  }
  if (flag == 0) {
    cout << "NO" << endl;
    return;
  } else {
    cout << "YES" << endl;
    cout << tot << endl;
    if (tot > 0) {
      for (int i = 1; i <= tot; i++) cout << c[i] << " ";
      cout << endl;
    }
  }
}
int main() {
  t = read();
  while (t--) {
    solve();
  }
  return 0;
}

```