#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
struct modint {
  int x;
  modint(int o = 0) { x = o; }
  modint &operator=(int o) { return x = o, *this; }
  modint &operator+=(modint o) {
    return x = x + o.x >= 1000000007 ? x + o.x - 1000000007 : x + o.x, *this;
  }
  modint &operator-=(modint o) {
    return x = x - o.x < 0 ? x - o.x + 1000000007 : x - o.x, *this;
  }
  modint &operator*=(modint o) { return x = 1ll * x * o.x % 1000000007, *this; }
  modint &operator^=(int b) {
    modint a = *this, c = 1;
    for (; b; b >>= 1, a *= a)
      if (b & 1) c *= a;
    return x = c.x, *this;
  }
  modint &operator/=(modint o) { return *this *= o ^= 1000000007 - 2; }
  modint &operator+=(int o) {
    return x = x + o >= 1000000007 ? x + o - 1000000007 : x + o, *this;
  }
  modint &operator-=(int o) {
    return x = x - o < 0 ? x - o + 1000000007 : x - o, *this;
  }
  modint &operator*=(int o) { return x = 1ll * x * o % 1000000007, *this; }
  modint &operator/=(int o) { return *this *= ((modint(o)) ^= 1000000007 - 2); }
  template <class I>
  friend modint operator+(modint a, I b) {
    return a += b;
  }
  template <class I>
  friend modint operator-(modint a, I b) {
    return a -= b;
  }
  template <class I>
  friend modint operator*(modint a, I b) {
    return a *= b;
  }
  template <class I>
  friend modint operator/(modint a, I b) {
    return a /= b;
  }
  friend modint operator^(modint a, int b) { return a ^= b; }
  friend bool operator==(modint a, int b) { return a.x == b; }
  friend bool operator!=(modint a, int b) { return a.x != b; }
  bool operator!() { return !x; }
  modint operator-() { return x ? 1000000007 - x : 0; }
};
int n, a[4000005], sum[4000005];
vector<int> res;
void work(int l, int r) {
  for (int i = r - 2; i >= l; i -= 2) res.push_back(i);
  for (int i = l + 2; i <= r - 2; i += 2) res.push_back(i);
}
void work() {
  n = read();
  res.clear();
  for (register int i = (1); i <= (n); ++i)
    a[i] = read(), sum[i] = sum[i - 1] ^ a[i];
  if (sum[n]) return puts("NO"), void();
  if (n & 1)
    work(1, n);
  else {
    bool ok = 0;
    for (register int i = (1); i <= (n); ++i)
      if (!sum[i] && (i & 1)) {
        ok = 1;
        work(1, i), work(i + 1, n);
        break;
      }
    if (!ok) return puts("NO"), void();
  }
  puts("YES");
  cout << res.size() << endl;
  for (auto v : res) cout << v << " ";
  puts("");
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}

### Prompt

Your challenge is to write 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;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
struct modint {
  int x;
  modint(int o = 0) { x = o; }
  modint &operator=(int o) { return x = o, *this; }
  modint &operator+=(modint o) {
    return x = x + o.x >= 1000000007 ? x + o.x - 1000000007 : x + o.x, *this;
  }
  modint &operator-=(modint o) {
    return x = x - o.x < 0 ? x - o.x + 1000000007 : x - o.x, *this;
  }
  modint &operator*=(modint o) { return x = 1ll * x * o.x % 1000000007, *this; }
  modint &operator^=(int b) {
    modint a = *this, c = 1;
    for (; b; b >>= 1, a *= a)
      if (b & 1) c *= a;
    return x = c.x, *this;
  }
  modint &operator/=(modint o) { return *this *= o ^= 1000000007 - 2; }
  modint &operator+=(int o) {
    return x = x + o >= 1000000007 ? x + o - 1000000007 : x + o, *this;
  }
  modint &operator-=(int o) {
    return x = x - o < 0 ? x - o + 1000000007 : x - o, *this;
  }
  modint &operator*=(int o) { return x = 1ll * x * o % 1000000007, *this; }
  modint &operator/=(int o) { return *this *= ((modint(o)) ^= 1000000007 - 2); }
  template <class I>
  friend modint operator+(modint a, I b) {
    return a += b;
  }
  template <class I>
  friend modint operator-(modint a, I b) {
    return a -= b;
  }
  template <class I>
  friend modint operator*(modint a, I b) {
    return a *= b;
  }
  template <class I>
  friend modint operator/(modint a, I b) {
    return a /= b;
  }
  friend modint operator^(modint a, int b) { return a ^= b; }
  friend bool operator==(modint a, int b) { return a.x == b; }
  friend bool operator!=(modint a, int b) { return a.x != b; }
  bool operator!() { return !x; }
  modint operator-() { return x ? 1000000007 - x : 0; }
};
int n, a[4000005], sum[4000005];
vector<int> res;
void work(int l, int r) {
  for (int i = r - 2; i >= l; i -= 2) res.push_back(i);
  for (int i = l + 2; i <= r - 2; i += 2) res.push_back(i);
}
void work() {
  n = read();
  res.clear();
  for (register int i = (1); i <= (n); ++i)
    a[i] = read(), sum[i] = sum[i - 1] ^ a[i];
  if (sum[n]) return puts("NO"), void();
  if (n & 1)
    work(1, n);
  else {
    bool ok = 0;
    for (register int i = (1); i <= (n); ++i)
      if (!sum[i] && (i & 1)) {
        ok = 1;
        work(1, i), work(i + 1, n);
        break;
      }
    if (!ok) return puts("NO"), void();
  }
  puts("YES");
  cout << res.size() << endl;
  for (auto v : res) cout << v << " ";
  puts("");
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}

```