#include <bits/stdc++.h>
using namespace std;
int _, n;
char a[100005], b[100005];
int ans[300005], tot = 0;
int ans2[300005], tot2 = 0;
int main() {
  for (scanf("%d", &_); _; --_) {
    tot = 0;
    tot2 = 0;
    scanf("%d%s%s", &n, a + 1, b + 1);
    int duana = 0, duanb = 0;
    int das[100005], now = 0, dbs[100005];
    int dasl[100005], dbsl[100005];
    for (int i = (1); i <= (n); ++i) {
      if (a[i] == '0') {
        if (!now) das[++duana] = i;
        ++now;
      } else if (now)
        dasl[duana] = i - 1, now = 0;
    }
    if (a[n] == '0') dasl[duana] = n;
    now = 0;
    for (int i = (1); i <= (n); ++i) {
      if (b[i] == '0') {
        if (!now) dbs[++duanb] = i;
        ++now;
      } else if (now)
        dbsl[duanb] = i - 1, now = 0;
    }
    if (b[n] == '0') dbsl[duanb] = n;
    for (int i = (1); i <= (duana); ++i) {
      if (das[i] - 1) ans[++tot] = das[i] - 1;
      ans[++tot] = dasl[i];
    }
    for (int i = (1); i <= (duanb); ++i) {
      if (dbs[i] - 1) ans2[++tot2] = dbs[i] - 1;
      ans2[++tot2] = dbsl[i];
    }
    printf("%d ", tot + tot2);
    for (int i = (1); i <= (tot); ++i) printf("%d ", ans[i]);
    for (int i = (tot2); i >= (1); --i) printf("%d ", ans2[i]);
    puts("");
  }
  return 0;
}

### 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>
using namespace std;
int _, n;
char a[100005], b[100005];
int ans[300005], tot = 0;
int ans2[300005], tot2 = 0;
int main() {
  for (scanf("%d", &_); _; --_) {
    tot = 0;
    tot2 = 0;
    scanf("%d%s%s", &n, a + 1, b + 1);
    int duana = 0, duanb = 0;
    int das[100005], now = 0, dbs[100005];
    int dasl[100005], dbsl[100005];
    for (int i = (1); i <= (n); ++i) {
      if (a[i] == '0') {
        if (!now) das[++duana] = i;
        ++now;
      } else if (now)
        dasl[duana] = i - 1, now = 0;
    }
    if (a[n] == '0') dasl[duana] = n;
    now = 0;
    for (int i = (1); i <= (n); ++i) {
      if (b[i] == '0') {
        if (!now) dbs[++duanb] = i;
        ++now;
      } else if (now)
        dbsl[duanb] = i - 1, now = 0;
    }
    if (b[n] == '0') dbsl[duanb] = n;
    for (int i = (1); i <= (duana); ++i) {
      if (das[i] - 1) ans[++tot] = das[i] - 1;
      ans[++tot] = dasl[i];
    }
    for (int i = (1); i <= (duanb); ++i) {
      if (dbs[i] - 1) ans2[++tot2] = dbs[i] - 1;
      ans2[++tot2] = dbsl[i];
    }
    printf("%d ", tot + tot2);
    for (int i = (1); i <= (tot); ++i) printf("%d ", ans[i]);
    for (int i = (tot2); i >= (1); --i) printf("%d ", ans2[i]);
    puts("");
  }
  return 0;
}

```