Problem: 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.

Solution (in JAVA):

Solution:
function not(c) {
   if (c === '1') {
      return '0';
   }
   return '1';
}

function solve() {
   var n = Number(read());
   var a = Array.from(read());
   var b = Array.from(read());
   var ans = [];

   function f(index) {
      var res = a.slice();
      for (var i = 0; i <= index; i++) {
         res[i] = not(a[index - i]);
      }
      a = res;
   }
   
   for (var i = n - 1; i >= 0; i--) {
      if (a[i] === b[i]) {
         continue;
      }
      if (a[0] !== b[i]) {
         ans.push(i + 1);
         f(i);
         continue;
      }
      var j = i - 1;
      while(a[j] !== b[i]) {
         j--;
      }
      ans.push(j + 1);
      f(j);
      ans.push(i + 1);
      f(i);
   }
   write(ans.length + ' ' + ans.join(' '));
}

function init() {
   var T;
   T = 1;
   T = Number(read());
   for (var t = 0; t < T; t++) {
      solve();
   }
}

function sortF(a, b) {
   if (a < b) {
      return -1;
   }
   if (a > b) {
      return 1;
   }
   return 0;
}

function readArray(f) {
   return read().split(' ').map(f);
}


var isNode = typeof console !== 'undefined';
var MEMORY = [];
var MEMORY_INDEX = 0;
if (isNode) {
   var fs = require('fs');
   var path = require('path');

   var inTextName;
   var outTextName;

   var needWriteFile = false;

   var writeResult = '';
   if (process.argv.length === 5) {
      inTextName = process.argv[3];
      outTextName = process.argv[4];
   }

   if (inTextName) {
      fs.readFile(path.join(__dirname, inTextName), 'utf8', (err, data) => {
         MEMORY = data.split('\r\n');
         init();
         if (needWriteFile) {
            fs.writeFile(path.join(__dirname, outTextName), writeResult, (err) => {
               err ? console.error(err) : console.log('good')
            });
         }
      });
   } else {
      var RL = require('readline').createInterface({
         input : process.stdin,
         output : process.stdout
      });

      RL.on('line', (line) => {
         MEMORY.push(line);
      });

      RL.on('close', init);
   }
} else {
   init();
}

function write(value) {
   if (needWriteFile) {
      writeResult += value + '\n';
      return;
   }
   isNode ? console.log(value) : print(value);
}
function read() {
   return isNode ? MEMORY[MEMORY_INDEX++] : readline();
}