This is the easy 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 3n 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≤ 1000) — 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 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), 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.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using ld = long double;
using pii = pair<int, int>;
namespace ImplicitTreap {
mt19937 rng(696969420);
struct node {
  int val;
  int count, prio;
  node *l, *r, *p;
  bool rev = false;
  string debug(int index) {
    return "(" + to_string(index) + "," + to_string(val) + ")";
  }
};
static int getCount(node* root) { return root ? root->count : 0; }
void prop(node* root) {
  if (!root) return;
  if (!root->rev) return;
  root->val ^= 1;
  swap(root->l, root->r);
  if (root->l) root->l->rev ^= 1;
  if (root->r) root->r->rev ^= 1;
  root->rev = false;
}
void update(node* root) {
  if (!root) return;
  root->p = NULL;
  if (root->l) root->l->p = root;
  if (root->r) root->r->p = root;
  root->count = 1 + getCount(root->l) + getCount(root->r);
}
void merge(node*& t, node* l, node* r) {
  prop(l), prop(r);
  if (!l || !r)
    t = l ? l : r;
  else if (l->prio > r->prio)
    merge(l->r, l->r, r), t = l;
  else
    merge(r->l, l, r->l), t = r;
  update(t);
}
void split(node* t, node*& l, node*& r, int index, int add = 0) {
  prop(t);
  if (!t) return void(l = r = NULL);
  int cur_index = add + getCount(t->l);
  if (index <= cur_index)
    split(t->l, l, t->l, index, add), r = t;
  else
    split(t->r, t->r, r, index, add + 1 + getCount(t->l)), l = t;
  update(t);
}
node* remove(node*& t, int index) {
  node *l, *r, *del;
  split(t, l, r, index);
  split(r, del, r, 1);
  merge(t, l, r);
  return del;
}
void insert(node*& t, node* item, int index) {
  node *l, *r;
  split(t, l, r, index);
  merge(l, l, item), merge(t, l, r);
}
node* insert(node*& root, int index, int val) {
  node* item = new node();
  item->val = val;
  item->count = 1;
  item->prio = rng();
  item->l = item->r = item->p = NULL;
  insert(root, item, index);
  return item;
}
node* get(node* t, int index, int add = 0) {
  int cur_index = add + getCount(t->l);
  if (index < cur_index) return get(t->l, index, add);
  if (index > cur_index) return get(t->r, index, add + 1 + getCount(t->l));
  return t;
}
int get_index(node* t) {
  int leq = getCount(t->l) + 1;
  while (t->p) {
    if (t->p->r == t) leq += getCount(t->p->l) + 1;
    t = t->p;
  }
  return leq - 1;
}
int getFirst(node* t) {
  prop(t);
  if (t->l) return getFirst(t->l);
  return t->val;
}
void print(node* t, int& index, bool isRoot) {
  if (!t) return;
  print(t->l, index, false);
  cerr << t->debug(index++) << ' ';
  print(t->r, index, false);
  if (isRoot) cout << endl;
}
void print(node* t) {
  int index = 0;
  print(t, index, true);
}
};  // namespace ImplicitTreap
vi ans;
vi go(vi& s, int k, vi& ans) {
  vi t = s;
  for (int i = 0; i < k + 1; i++) t[i] = s[k - i] ^ 1;
  ans.push_back(k + 1);
  return t;
}
void solve() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  vi a(n), b(n);
  vi ans;
  ImplicitTreap::node* root = NULL;
  for (int i = 0; i < n; i++)
    a[i] = s[i] == '1', ImplicitTreap::insert(root, i, a[i]);
  for (int i = 0; i < n; i++) b[i] = t[i] == '1';
  for (int i = n - 1; i >= 0; i--) {
    int first = ImplicitTreap::getFirst(root);
    ImplicitTreap::node *l, *r;
    ImplicitTreap::split(root, l, r, i + 1);
    l->rev = true;
    ImplicitTreap::merge(root, l, r);
    if (first != b[i])
      ans.push_back(i + 1);
    else {
      ans.push_back(1);
      ans.push_back(i + 1);
    }
  }
  cout << ans.size() << " ";
  for (int i : ans) cout << i << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
