Problem: You have two strings a and b of equal even length n consisting of characters 0 and 1.

We're in the endgame now. To finally make the universe perfectly balanced, you need to make strings a and b equal.

In one step, you can choose any prefix of a of even length and reverse it. Formally, if a = a_1 a_2 … a_n, you can choose a positive even integer p ≤ n and set a to a_p a_{p-1} … a_1 a_{p+1} a_{p+2} … a_n.

Find a way to make a equal to b using at most n + 1 reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized.

Input

The first line contains a single integer t (1 ≤ t ≤ 2000), denoting the number of test cases.

Each test case consists of two lines. The first line contains a string a of length n, and the second line contains a string b of the same length (2 ≤ n ≤ 4000; n mod 2 = 0). Both strings consist of characters 0 and 1.

The sum of n over all t test cases doesn't exceed 4000.

Output

For each test case, if it's impossible to make a equal to b in at most n + 1 reversals, output a single integer -1.

Otherwise, output an integer k (0 ≤ k ≤ n + 1), denoting the number of reversals in your sequence of steps, followed by k even integers p_1, p_2, …, p_k (2 ≤ p_i ≤ n; p_i mod 2 = 0), denoting the lengths of prefixes of a to be reversed, in chronological order.

Note that k doesn't have to be minimized. If there are many solutions, output any of them.

Example

Input


4
0100011011
1101011000
10101010
10101010
0011
1001
100011
110010


Output


3
6 4 10
0

-1
7
2 6 2 6 2 2 6

Note

In the first test case, string a changes as follows: 

  * after the first reversal: 1000101011; 
  * after the second reversal: 0001101011; 
  * after the third reversal: 1101011000. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 4e3 + 10;
int s[N], t[N];
std::vector<int> res;
inline void Rev(int n) {
  if (!n) return;
  std::reverse(s + 1, s + n + 1);
  res.emplace_back(n);
}
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  int T;
  std::cin >> T;
  for (int cas = 1; cas <= T; ++cas) {
    res.clear();
    static char S[N], T[N];
    std::cin >> S + 1 >> T + 1;
    int n = strlen(S + 1);
    for (int i = 1; i <= n; ++i) s[i] = S[i] - '0', t[i] = T[i] - '0';
    int a[4] = {0, 0, 0, 0}, b[4] = {0, 0, 0, 0};
    for (int i = 1; i <= n; i += 2)
      a[s[i] * 2 + s[i + 1]]++, b[t[i] * 2 + t[i + 1]]++;
    if (a[0] != b[0] || a[3] != b[3]) {
      std::cout << -1 << "\n";
      continue;
    }
    int p = 0;
    if (a[1] != b[2]) {
      int bp = a[1] - a[2] + b[1] - b[2] >> 1, cnt = 0;
      if (abs(a[1] - a[2]) >= abs(b[1] - b[2]))
        for (int i = 1; i <= n; i += 2) {
          if (!s[i] && s[i + 1]) ++cnt;
          if (s[i] && !s[i + 1]) --cnt;
          if (cnt == bp) {
            Rev(i + 1);
            break;
          }
        }
      else
        for (int i = 1; i <= n; i += 2) {
          if (!t[i] && t[i + 1]) ++cnt;
          if (t[i] && !t[i + 1]) --cnt;
          if (cnt == bp) {
            std::reverse(t + 1, t + i + 2);
            p = i + 1;
            break;
          }
        }
    }
    for (int i = 1; i <= n; i += 2) {
      int k = i;
      while (s[k + 1] != t[n - i] || s[k] != t[n - i + 1]) k += 2;
      Rev(k - 1), Rev(k + 1);
    }
    if (p) Rev(p);
    std::cout << res.size() << "\n";
    for (int i = 0; i < res.size(); ++i)
      std::cout << res[i] << " \n"[i == res.size() - 1];
  }
  return 0;
}