#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
inline long long gcd(long long a, long long b) {
  return !b ? a : gcd(b, a % b);
}
inline long long q_pow(long long a, long long x = mod - 2) {
  long long ans = 1, tmp = a;
  while (x) {
    if (x & 1) (ans *= tmp) %= mod;
    (tmp *= tmp) %= mod;
    x >>= 1;
  }
  return ans;
}
template <typename T>
inline void re(T &N) {
  int f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  N = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
  N *= f;
}
template <class T, class... T_>
inline void re(T &x, T_ &...y) {
  re(x), re(y...);
}
int m, n, t = 1, st, en;
int a[N], b[N];
char s[N];
int main() {
  re(t);
  while (t--) {
    re(n);
    int now = 0;
    for (int i = 1; i <= n; i++) re(a[i]), now ^= a[i];
    if (now) {
      puts("NO");
      continue;
    }
    vector<int> ans;
    int ok = 1;
    for (int i = 1; i <= n && ok; i++)
      if (a[i]) {
        int j = i;
        while (j <= n && a[j]) j++;
        int cnt = j - i;
        while (cnt & 1) {
          while (j <= n && !a[j]) {
            if (j == n) {
              ok = 0;
              break;
            }
            if (!a[j + 1])
              ans.push_back(j - 1), a[j] = a[j + 1] = 1, j += 2;
            else {
              ans.push_back(j - 1), a[j - 1] = a[j] = a[j + 1] = 0;
              j--;
              break;
            }
          }
          if (!ok) break;
          if (a[j]) {
            while (j <= n && a[j]) j++;
            cnt = j - i;
          } else
            break;
        }
        if (i > 1 && !a[i - 1]) {
          for (int k = i; k < j; k += 2) {
            ans.push_back(k - 1);
            a[k] = a[k + 1] = 0;
          }
        } else if (j <= n && !a[j]) {
          for (int k = j - 2; k >= i; k -= 2) {
            ans.push_back(k);
            a[k] = a[k + 1] = 0;
          }
        } else
          ok = 0;
        i = j - 1;
      }
    if (!ok)
      puts("NO");
    else {
      puts("YES");
      printf("%d\n", (int)ans.size());
      if (ans.size()) {
        for (int i : ans) printf("%d ", i);
        puts("");
      }
    }
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 1e6 + 10;
const long long mod = 1e9 + 7;
inline long long gcd(long long a, long long b) {
  return !b ? a : gcd(b, a % b);
}
inline long long q_pow(long long a, long long x = mod - 2) {
  long long ans = 1, tmp = a;
  while (x) {
    if (x & 1) (ans *= tmp) %= mod;
    (tmp *= tmp) %= mod;
    x >>= 1;
  }
  return ans;
}
template <typename T>
inline void re(T &N) {
  int f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  N = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
  N *= f;
}
template <class T, class... T_>
inline void re(T &x, T_ &...y) {
  re(x), re(y...);
}
int m, n, t = 1, st, en;
int a[N], b[N];
char s[N];
int main() {
  re(t);
  while (t--) {
    re(n);
    int now = 0;
    for (int i = 1; i <= n; i++) re(a[i]), now ^= a[i];
    if (now) {
      puts("NO");
      continue;
    }
    vector<int> ans;
    int ok = 1;
    for (int i = 1; i <= n && ok; i++)
      if (a[i]) {
        int j = i;
        while (j <= n && a[j]) j++;
        int cnt = j - i;
        while (cnt & 1) {
          while (j <= n && !a[j]) {
            if (j == n) {
              ok = 0;
              break;
            }
            if (!a[j + 1])
              ans.push_back(j - 1), a[j] = a[j + 1] = 1, j += 2;
            else {
              ans.push_back(j - 1), a[j - 1] = a[j] = a[j + 1] = 0;
              j--;
              break;
            }
          }
          if (!ok) break;
          if (a[j]) {
            while (j <= n && a[j]) j++;
            cnt = j - i;
          } else
            break;
        }
        if (i > 1 && !a[i - 1]) {
          for (int k = i; k < j; k += 2) {
            ans.push_back(k - 1);
            a[k] = a[k + 1] = 0;
          }
        } else if (j <= n && !a[j]) {
          for (int k = j - 2; k >= i; k -= 2) {
            ans.push_back(k);
            a[k] = a[k + 1] = 0;
          }
        } else
          ok = 0;
        i = j - 1;
      }
    if (!ok)
      puts("NO");
    else {
      puts("YES");
      printf("%d\n", (int)ans.size());
      if (ans.size()) {
        for (int i : ans) printf("%d ", i);
        puts("");
      }
    }
  }
}

```