#include <bits/stdc++.h>
using namespace std;
const int N = 1000007, inf = 0x3f3f3f3f;
void invert(string& s, int n) {
  for (int i = 0; i < n; i++) {
    s[i] = 1 - s[i];
  }
}
void revert(string& s, int n) {
  for (int l = 0, r = n - 1; l < r; ++l, --r) {
    swap(s[l], s[r]);
  }
}
bool check(vector<int>& op, string s, string t) {
  for (int x : op) {
    invert(s, x);
    revert(s, x);
  }
  return s == t;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  auto _solve = []() {
    int n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    for (char& c : s) c -= '0';
    for (char& c : t) c -= '0';
    string auxs = s, auxt = t;
    vector<int> ans;
    bool rev = false;
    int l = 0, r = n - 1;
    int pos = n - 1;
    while (l < r) {
      if (rev) {
        if (t[pos] == (s[l] ^ rev)) {
          l++;
        } else {
          if (s[l] != s[r]) {
            ans.push_back(1);
            s[r] = 1 - s[r];
          }
          ans.push_back(r - l + 1);
          rev ^= 1;
          r--;
        }
      } else {
        if (t[pos] == (s[r] ^ rev)) {
          r--;
        } else {
          if (s[l] != s[r]) {
            ans.push_back(1);
            s[l] = 1 - s[l];
          }
          ans.push_back(r - l + 1);
          rev ^= 1;
          l++;
        }
      }
      pos--;
    }
    if (l == r) {
      if ((s[l] ^ rev) != (t[pos])) ans.push_back(1);
    }
    cout << ans.size() << ' ';
    for (int x : ans) cout << x << ' ';
    cout << '\n';
  };
  int t;
  cin >> t;
  while (t--) _solve();
}

### Prompt

In CPP, your task is to solve 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 int N = 1000007, inf = 0x3f3f3f3f;
void invert(string& s, int n) {
  for (int i = 0; i < n; i++) {
    s[i] = 1 - s[i];
  }
}
void revert(string& s, int n) {
  for (int l = 0, r = n - 1; l < r; ++l, --r) {
    swap(s[l], s[r]);
  }
}
bool check(vector<int>& op, string s, string t) {
  for (int x : op) {
    invert(s, x);
    revert(s, x);
  }
  return s == t;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  auto _solve = []() {
    int n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    for (char& c : s) c -= '0';
    for (char& c : t) c -= '0';
    string auxs = s, auxt = t;
    vector<int> ans;
    bool rev = false;
    int l = 0, r = n - 1;
    int pos = n - 1;
    while (l < r) {
      if (rev) {
        if (t[pos] == (s[l] ^ rev)) {
          l++;
        } else {
          if (s[l] != s[r]) {
            ans.push_back(1);
            s[r] = 1 - s[r];
          }
          ans.push_back(r - l + 1);
          rev ^= 1;
          r--;
        }
      } else {
        if (t[pos] == (s[r] ^ rev)) {
          r--;
        } else {
          if (s[l] != s[r]) {
            ans.push_back(1);
            s[l] = 1 - s[l];
          }
          ans.push_back(r - l + 1);
          rev ^= 1;
          l++;
        }
      }
      pos--;
    }
    if (l == r) {
      if ((s[l] ^ rev) != (t[pos])) ans.push_back(1);
    }
    cout << ans.size() << ' ';
    for (int x : ans) cout << x << ' ';
    cout << '\n';
  };
  int t;
  cin >> t;
  while (t--) _solve();
}

```