#include <bits/stdc++.h>
const long double PI = acos(-1);
using namespace std;
inline long long mod(long long n, long long m = 1000000007) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
struct BIT {
  vector<long long> bit;
  BIT() {}
  long long n;
  BIT(long long n) {
    this->n = n;
    bit.resize(n + 1);
  }
  void update(long long idx, long long val) {
    for (long long i = idx; i <= n; i += i & -i) {
      bit[i] += val;
    }
  }
  long long prefix_query(long long idx) {
    long long ans = 0;
    for (long long i = idx; i > 0; i -= i & -i) {
      ans += bit[i];
    }
    return ans;
  }
  long long query(long long l, long long r) {
    return prefix_query(r) - prefix_query(l - 1);
  }
  long long kth(long long k) {
    long long cur = 0;
    long long acc = 0;
    for (long long i = 19; i >= 0; i--) {
      if (cur + (1 << i) > n) continue;
      if (acc + bit[cur + (1 << i)] < k) {
        cur += (1 << i);
        acc += bit[cur];
      }
    }
    return ++cur;
  }
};
long long exp(long long n, long long k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  long long ax = exp(n, k / 2);
  ax = mod(ax * ax);
  if (k % 2) ax = mod(ax * n);
  return ax;
}
long long gcd(long long a, long long b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
void solution() {
  long long n;
  string s, t;
  cin >> n >> s >> t;
  long long l = 0, r = n - 1;
  bool rev = false;
  vector<long long> ans;
  for (long long i = n - 1; i >= 0; i--) {
    if (!rev) {
      if (s[l] == t[i]) ans.push_back(0);
      ans.push_back(i);
      l++;
      rev = !rev;
    } else {
      if (s[r] != t[i]) ans.push_back(0);
      ans.push_back(i);
      r--;
      rev = !rev;
    }
  }
  cout << ans.size();
  for (long long x : ans) cout << " " << x + 1;
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long t = 1;
  cin >> t;
  while (t--) solution();
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
const long double PI = acos(-1);
using namespace std;
inline long long mod(long long n, long long m = 1000000007) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
struct BIT {
  vector<long long> bit;
  BIT() {}
  long long n;
  BIT(long long n) {
    this->n = n;
    bit.resize(n + 1);
  }
  void update(long long idx, long long val) {
    for (long long i = idx; i <= n; i += i & -i) {
      bit[i] += val;
    }
  }
  long long prefix_query(long long idx) {
    long long ans = 0;
    for (long long i = idx; i > 0; i -= i & -i) {
      ans += bit[i];
    }
    return ans;
  }
  long long query(long long l, long long r) {
    return prefix_query(r) - prefix_query(l - 1);
  }
  long long kth(long long k) {
    long long cur = 0;
    long long acc = 0;
    for (long long i = 19; i >= 0; i--) {
      if (cur + (1 << i) > n) continue;
      if (acc + bit[cur + (1 << i)] < k) {
        cur += (1 << i);
        acc += bit[cur];
      }
    }
    return ++cur;
  }
};
long long exp(long long n, long long k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  long long ax = exp(n, k / 2);
  ax = mod(ax * ax);
  if (k % 2) ax = mod(ax * n);
  return ax;
}
long long gcd(long long a, long long b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
void solution() {
  long long n;
  string s, t;
  cin >> n >> s >> t;
  long long l = 0, r = n - 1;
  bool rev = false;
  vector<long long> ans;
  for (long long i = n - 1; i >= 0; i--) {
    if (!rev) {
      if (s[l] == t[i]) ans.push_back(0);
      ans.push_back(i);
      l++;
      rev = !rev;
    } else {
      if (s[r] != t[i]) ans.push_back(0);
      ans.push_back(i);
      r--;
      rev = !rev;
    }
  }
  cout << ans.size();
  for (long long x : ans) cout << " " << x + 1;
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long t = 1;
  cin >> t;
  while (t--) solution();
}

```