#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 1e6 + 5;
const long long MOD = 1000000007;
const long long MOD2 = 2010405347;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 315;
inline long long pw(long long x, long long y, long long md = MOD) {
  long long ret = 1;
  x %= md;
  while (y) {
    if (y & 1) ret = ret * x % md;
    x = x * x % md, y >>= 1;
  }
  return ret;
}
inline void add(int &a, int b, int md = MOD) {
  a = a + b >= md ? a + b - md : a + b;
}
long long tc, n, x[MAX], y[MAX], cnt, hit;
vector<int> ans;
void solve(int le, int ri) {
  cnt = x[le] + x[le + 1];
  int tmp = 0;
  for (int i = (le); i <= (ri - 2); ++i) {
    cnt += x[i + 2];
    if (cnt % 2 == 0 && cnt) {
      tmp = 0;
      for (int k = (0); k <= (2); ++k) tmp += x[i + k];
      if (tmp == 2) {
        ans.push_back(i);
        for (int k = (0); k <= (2); ++k) x[i + k] = 0;
        for (int j = (i); j >= (le); --j) {
          tmp = 0;
          for (int k = (0); k <= (2); ++k) tmp += x[j + k];
          if (tmp == 2) {
            ans.push_back(j);
            for (int k = (0); k <= (2); ++k) x[j + k] = 0;
          }
        }
        for (int j = (i); j <= (ri - 2); ++j) {
          tmp = 0;
          for (int k = (0); k <= (2); ++k) tmp += x[j + k];
          if (tmp == 2) {
            ans.push_back(j);
            for (int k = (0); k <= (2); ++k) x[j + k] = 0;
          }
        }
        return;
      }
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  for (int ntc = (1); ntc <= (tc); ++ntc) {
    cin >> n;
    cnt = 0;
    for (int i = (1); i <= (n); ++i) {
      cin >> x[i];
      y[i] = x[i];
      cnt += x[i];
    }
    if ((cnt & 1) || cnt == n) {
      cout << "NO\n";
      continue;
    }
    ans.clear();
    hit = cnt;
    cnt = 0;
    bool st = 0;
    for (int i = (1); i <= (n); ++i)
      cnt += x[i], st |= (cnt % 2 == 0) && (i & 1);
    cnt = 0;
    for (int i = (1); i <= (n); ++i) {
      cnt += x[i];
      if (cnt && (cnt % 2 == 0) && (i & 1)) {
        hit -= cnt;
        for (int j = (1); j <= (i - 2); ++j)
          if (x[j]) {
            cnt = 0;
            for (int k = (0); k <= (2); ++k) cnt += x[j + k];
            if (cnt == 1) {
              for (int k = (0); k <= (2); ++k) x[j + k] = 1;
              ans.push_back(j);
            }
          }
        solve(1, i);
        break;
      }
    }
    cnt = 0;
    for (int i = (n); i >= (1); --i) {
      cnt += x[i];
      if (cnt == hit && (i % 2 == n % 2)) {
        for (int j = (i); j <= (n - 2); ++j)
          if (x[j]) {
            cnt = 0;
            for (int k = (0); k <= (2); ++k) cnt += x[j + k];
            if (cnt == 1) {
              for (int k = (0); k <= (2); ++k) x[j + k] = 1;
              ans.push_back(j);
            }
          }
        solve(i, n);
        break;
      }
    }
    cnt = 0;
    for (int i = (1); i <= (n); ++i) {
      cnt += x[i];
    }
    if (cnt) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      assert(ans.size() <= n);
      cout << ans.size() << '\n';
      for (int i : ans) cout << i << " ";
      cout << '\n';
    }
  }
  return 0;
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC target("tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 1e6 + 5;
const long long MOD = 1000000007;
const long long MOD2 = 2010405347;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 315;
inline long long pw(long long x, long long y, long long md = MOD) {
  long long ret = 1;
  x %= md;
  while (y) {
    if (y & 1) ret = ret * x % md;
    x = x * x % md, y >>= 1;
  }
  return ret;
}
inline void add(int &a, int b, int md = MOD) {
  a = a + b >= md ? a + b - md : a + b;
}
long long tc, n, x[MAX], y[MAX], cnt, hit;
vector<int> ans;
void solve(int le, int ri) {
  cnt = x[le] + x[le + 1];
  int tmp = 0;
  for (int i = (le); i <= (ri - 2); ++i) {
    cnt += x[i + 2];
    if (cnt % 2 == 0 && cnt) {
      tmp = 0;
      for (int k = (0); k <= (2); ++k) tmp += x[i + k];
      if (tmp == 2) {
        ans.push_back(i);
        for (int k = (0); k <= (2); ++k) x[i + k] = 0;
        for (int j = (i); j >= (le); --j) {
          tmp = 0;
          for (int k = (0); k <= (2); ++k) tmp += x[j + k];
          if (tmp == 2) {
            ans.push_back(j);
            for (int k = (0); k <= (2); ++k) x[j + k] = 0;
          }
        }
        for (int j = (i); j <= (ri - 2); ++j) {
          tmp = 0;
          for (int k = (0); k <= (2); ++k) tmp += x[j + k];
          if (tmp == 2) {
            ans.push_back(j);
            for (int k = (0); k <= (2); ++k) x[j + k] = 0;
          }
        }
        return;
      }
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  for (int ntc = (1); ntc <= (tc); ++ntc) {
    cin >> n;
    cnt = 0;
    for (int i = (1); i <= (n); ++i) {
      cin >> x[i];
      y[i] = x[i];
      cnt += x[i];
    }
    if ((cnt & 1) || cnt == n) {
      cout << "NO\n";
      continue;
    }
    ans.clear();
    hit = cnt;
    cnt = 0;
    bool st = 0;
    for (int i = (1); i <= (n); ++i)
      cnt += x[i], st |= (cnt % 2 == 0) && (i & 1);
    cnt = 0;
    for (int i = (1); i <= (n); ++i) {
      cnt += x[i];
      if (cnt && (cnt % 2 == 0) && (i & 1)) {
        hit -= cnt;
        for (int j = (1); j <= (i - 2); ++j)
          if (x[j]) {
            cnt = 0;
            for (int k = (0); k <= (2); ++k) cnt += x[j + k];
            if (cnt == 1) {
              for (int k = (0); k <= (2); ++k) x[j + k] = 1;
              ans.push_back(j);
            }
          }
        solve(1, i);
        break;
      }
    }
    cnt = 0;
    for (int i = (n); i >= (1); --i) {
      cnt += x[i];
      if (cnt == hit && (i % 2 == n % 2)) {
        for (int j = (i); j <= (n - 2); ++j)
          if (x[j]) {
            cnt = 0;
            for (int k = (0); k <= (2); ++k) cnt += x[j + k];
            if (cnt == 1) {
              for (int k = (0); k <= (2); ++k) x[j + k] = 1;
              ans.push_back(j);
            }
          }
        solve(i, n);
        break;
      }
    }
    cnt = 0;
    for (int i = (1); i <= (n); ++i) {
      cnt += x[i];
    }
    if (cnt) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      assert(ans.size() <= n);
      cout << ans.size() << '\n';
      for (int i : ans) cout << i << " ";
      cout << '\n';
    }
  }
  return 0;
}

```