#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
inline long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
inline long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
inline long long int add(long long int a, long long int b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}
inline long long int multi(long long int a, long long int b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
inline long long int sub(long long int a, long long int b) {
  a %= MOD;
  b %= MOD;
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline long long int power(long long int a, long long int b) {
  a %= MOD;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = multi(res, a);
    }
    a = multi(a, a);
    b >>= 1;
  }
  return res;
}
bool isPrime(long long int x) {
  if (x <= 4 || x % 2 == 0 || x % 3 == 0) {
    return x == 2 || x == 3;
  }
  for (long long int i = 5; i * i <= x; i += 6) {
    if (x % i == 0 || x % (i + 2) == 0) {
      return 0;
    }
  }
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long int x = 0, f = 0;
    long long int start = 0, end = n - 1;
    vector<long long int> res;
    for (long long int i = n - 1; i >= 0; i--) {
      long long int k = x == start ? end : start;
      if (k == x) {
        if (f && a[x] == b[i])
          res.push_back(1);
        else if (!f && a[x] != b[i])
          res.push_back(1);
        break;
      }
      if (f == 1) {
        if (a[k] == '1')
          a[k] = '0';
        else
          a[k] = '1';
        if (a[x] == '1')
          a[x] = '0';
        else
          a[x] = '1';
      }
      if (a[k] != b[i]) {
        if (a[x] == b[i]) res.push_back(1);
        res.push_back(end - start + 1);
        if (f == 1) {
          if (a[k] == '1')
            a[k] = '0';
          else
            a[k] = '1';
          if (a[x] == '1')
            a[x] = '0';
          else
            a[x] = '1';
        }
        if (x == start) {
          x = end;
          start++;
        } else {
          x = start;
          end--;
        }
        f ^= 1;
      } else {
        if (f == 1) {
          if (a[k] == '1')
            a[k] = '0';
          else
            a[k] = '1';
          if (a[x] == '1')
            a[x] = '0';
          else
            a[x] = '1';
        }
        if (x == start)
          end--;
        else
          start++;
      }
    }
    cout << res.size() << " ";
    for (long long int i : res) cout << i << " ";
    cout << endl;
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

This is the hard version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.

For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.

Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the length of the binary strings.

The next two lines contain two binary strings a and b of length n.

It is guaranteed that the sum of n across all test cases does not exceed 10^5.

Output

For each test case, output an integer k (0≤ k≤ 2n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


Output


3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
inline long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
inline long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
inline long long int add(long long int a, long long int b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}
inline long long int multi(long long int a, long long int b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
inline long long int sub(long long int a, long long int b) {
  a %= MOD;
  b %= MOD;
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline long long int power(long long int a, long long int b) {
  a %= MOD;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = multi(res, a);
    }
    a = multi(a, a);
    b >>= 1;
  }
  return res;
}
bool isPrime(long long int x) {
  if (x <= 4 || x % 2 == 0 || x % 3 == 0) {
    return x == 2 || x == 3;
  }
  for (long long int i = 5; i * i <= x; i += 6) {
    if (x % i == 0 || x % (i + 2) == 0) {
      return 0;
    }
  }
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long int x = 0, f = 0;
    long long int start = 0, end = n - 1;
    vector<long long int> res;
    for (long long int i = n - 1; i >= 0; i--) {
      long long int k = x == start ? end : start;
      if (k == x) {
        if (f && a[x] == b[i])
          res.push_back(1);
        else if (!f && a[x] != b[i])
          res.push_back(1);
        break;
      }
      if (f == 1) {
        if (a[k] == '1')
          a[k] = '0';
        else
          a[k] = '1';
        if (a[x] == '1')
          a[x] = '0';
        else
          a[x] = '1';
      }
      if (a[k] != b[i]) {
        if (a[x] == b[i]) res.push_back(1);
        res.push_back(end - start + 1);
        if (f == 1) {
          if (a[k] == '1')
            a[k] = '0';
          else
            a[k] = '1';
          if (a[x] == '1')
            a[x] = '0';
          else
            a[x] = '1';
        }
        if (x == start) {
          x = end;
          start++;
        } else {
          x = start;
          end--;
        }
        f ^= 1;
      } else {
        if (f == 1) {
          if (a[k] == '1')
            a[k] = '0';
          else
            a[k] = '1';
          if (a[x] == '1')
            a[x] = '0';
          else
            a[x] = '1';
        }
        if (x == start)
          end--;
        else
          start++;
      }
    }
    cout << res.size() << " ";
    for (long long int i : res) cout << i << " ";
    cout << endl;
  }
}

```