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;
const int N = 1e5 + 10;
char a[N], b[N];
int n, T;
int ans[N * 2], ct, rev;
int rt, ch[N][2], fa[N], tg[N], col[N], sz[N];
int get(int x) { return ch[fa[x]][1] == x; }
void pushdown(int x) {
  swap(ch[x][0], ch[x][1]), col[ch[x][0]] ^= 1, col[ch[x][1]] ^= 1,
      tg[ch[x][0]] ^= 1, tg[ch[x][1]] ^= 1;
  tg[x] = 0;
}
int s(int x, int k) {
  if (tg[x]) pushdown(x);
  if (sz[ch[x][0]] >= k) return s(ch[x][0], k);
  if (sz[ch[x][0]] + 1 == k) return x;
  return s(ch[x][1], k - sz[ch[x][0]] - 1);
}
char sc(int k) { return col[s(rt, k)] + '0'; }
void updata(int x) { sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1; }
void rotate(int x) {
  int k = get(x), y = fa[x];
  ch[y][k] = ch[x][k ^ 1];
  fa[ch[y][k]] = y;
  if (fa[y]) ch[fa[y]][ch[fa[y]][1] == y] = x;
  fa[x] = fa[y];
  fa[y] = x;
  ch[x][k ^ 1] = y;
  updata(y);
  updata(x);
}
int q[N], cnt;
void splay(int x) {
  int y;
  q[cnt = 1] = x;
  for (int i = x; fa[i]; i = fa[i]) q[++cnt] = fa[i];
  for (int i = cnt; i >= 1; i--)
    if (tg[q[i]]) pushdown(q[i]);
  while (fa[x]) {
    y = fa[x];
    if (fa[y]) rotate(get(x) == get(y) ? y : x);
    rotate(x);
  }
}
void work(int k) {
  if (k == n) {
    col[rt] ^= 1;
    tg[rt] ^= 1;
    return;
  }
  int y = s(rt, k + 1);
  splay(y);
  rt = y;
  col[ch[y][0]] ^= 1, tg[ch[y][0]] ^= 1;
}
void clr() {
  for (int i = 1; i <= n; i++) fa[i] = tg[i] = sz[i] = ch[i][0] = ch[i][1] = 0;
}
int ins(int l, int r) {
  if (l > r) return 0;
  int mid = l + r >> 1;
  col[mid] = a[mid] - '0';
  ch[mid][0] = ins(l, mid - 1);
  fa[ch[mid][0]] = mid;
  ch[mid][1] = ins(mid + 1, r);
  fa[ch[mid][1]] = mid;
  sz[mid] = sz[ch[mid][0]] + sz[ch[mid][1]] + 1;
  return mid;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    ct = rev = 0;
    clr();
    char c;
    int aaa = 1;
    scanf("%d%s%s", &n, a + 1, b + 1);
    rt = ins(1, n);
    for (int i = n; i; i--)
      if ((c = sc(i)) != b[i]) {
        if (sc(1) != c) ans[++ct] = 1, work(1);
        ans[++ct] = i;
        work(i);
      }
    sc(1);
    printf("%d", ct);
    for (int i = 1; i <= ct; i++) printf(" %d", ans[i]);
    puts("");
  }
}
