#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
inline long long Read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
int n;
char s1[100005], s2[100005];
int ans[200005], top;
struct List {
  int l, r;
  char v;
} nd[100005];
bool rev;
int frt() { return rev ? nd[n + 1].l : nd[0].r; }
int bac() { return rev ? nd[0].r : nd[n + 1].l; }
char val(int p) { return rev == nd[p].v - '0' ? '0' : '1'; }
void del(int p) {
  nd[nd[p].l].r = nd[p].r;
  nd[nd[p].r].l = nd[p].l;
}
void solve(int t) {
  n = read();
  scanf("%s%s", s1 + 1, s2 + 1);
  for (int i = 1; i <= n; i++) {
    nd[i].l = i - 1, nd[i].r = i + 1;
    nd[i].v = s1[i];
  }
  nd[0].r = 1, nd[n + 1].l = n;
  rev = 0, top = 0;
  for (int i = n; i >= 1; i--) {
    int b = bac();
    if (val(b) == s2[i]) {
      del(b);
      continue;
    }
    int f = frt();
    if (val(f) == s2[i]) ans[++top] = 1;
    rev ^= 1;
    ans[++top] = i;
    del(f);
  }
  printf("%d ", top);
  for (int i = 1; i <= top; i++) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  int t = read();
  for (int i = 1; i <= t; i++) solve(i);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
inline long long Read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
int n;
char s1[100005], s2[100005];
int ans[200005], top;
struct List {
  int l, r;
  char v;
} nd[100005];
bool rev;
int frt() { return rev ? nd[n + 1].l : nd[0].r; }
int bac() { return rev ? nd[0].r : nd[n + 1].l; }
char val(int p) { return rev == nd[p].v - '0' ? '0' : '1'; }
void del(int p) {
  nd[nd[p].l].r = nd[p].r;
  nd[nd[p].r].l = nd[p].l;
}
void solve(int t) {
  n = read();
  scanf("%s%s", s1 + 1, s2 + 1);
  for (int i = 1; i <= n; i++) {
    nd[i].l = i - 1, nd[i].r = i + 1;
    nd[i].v = s1[i];
  }
  nd[0].r = 1, nd[n + 1].l = n;
  rev = 0, top = 0;
  for (int i = n; i >= 1; i--) {
    int b = bac();
    if (val(b) == s2[i]) {
      del(b);
      continue;
    }
    int f = frt();
    if (val(f) == s2[i]) ans[++top] = 1;
    rev ^= 1;
    ans[++top] = i;
    del(f);
  }
  printf("%d ", top);
  for (int i = 1; i <= top; i++) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  int t = read();
  for (int i = 1; i <= t; i++) solve(i);
  return 0;
}

```