You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t. 

You can perform the following operation on s any number of times to achieve it — 

  * Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position. 

For example, on rotating the substring [2,4] , string "abcde" becomes "adbce". 




A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Find the minimum number of operations required to convert s to t, or determine that it's impossible.

Input

The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings. 

The second and the third lines contain strings s and t respectively.

The sum of n over all the test cases does not exceed 2000.

Output

For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.

Example

Input


6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc


Output


0
1
1
2
1
-1

Note

For the 1-st test case, since s and t are equal, you don't need to apply any operation.

For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.

For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.

For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.

For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.

For the 6-th test case, it is not possible to convert string s to t.
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
int w;
int dp[2005][2005];
string s, t;
int quick_s[2005][30], quick_ts[2005][30];
int solve(int i, int j) {
  if (i == 0)
    return 0;
  else {
    int cur;
    if (dp[i - 1][j] == INF)
      cur = solve(i - 1, j) + 1;
    else
      cur = dp[i - 1][j] + 1;
    if (s.at(i) == t.at(j)) {
      if (dp[i - 1][j - 1] == INF)
        dp[i][j] = solve(i - 1, j - 1);
      else
        dp[i][j] = dp[i - 1][j - 1];
      return dp[i][j];
    } else if (j >= i) {
      int prob = t.at(j) - 'a';
      if (quick_ts[j][prob] > quick_s[i][prob]) {
        if (dp[i][j - 1] == INF)
          dp[i][j] = solve(i, j - 1);
        else
          dp[i][j] = dp[i][j - 1];
      } else {
        if (dp[i - 1][j] == INF)
          dp[i][j] = solve(i - 1, j) + 1;
        else
          dp[i][j] = dp[i - 1][j] + 1;
      }
      dp[i][j] = min(cur, dp[i][j]);
      return dp[i][j];
    } else {
      return 0;
    }
  }
}
int main() {
  cin >> w;
  while (w--) {
    int n, i, j;
    cin >> n >> s >> t;
    for (i = 0; i <= n; i++) {
      dp[0][i] = 0;
    }
    for (i = 0; i < 30; i++) {
      quick_s[0][i] = 0;
      quick_ts[0][i] = 0;
    }
    quick_s[0][s.at(0) - 'a'] = 1;
    quick_ts[0][t.at(0) - 'a'] = 1;
    for (i = 1; i < n; i++) {
      for (j = 0; j < 30; j++) {
        if (j == s.at(i) - 'a')
          quick_s[i][j] = quick_s[i - 1][j] + 1;
        else
          quick_s[i][j] = quick_s[i - 1][j];
        if (j == t.at(i) - 'a')
          quick_ts[i][j] = quick_ts[i - 1][j] + 1;
        else
          quick_ts[i][j] = quick_ts[i - 1][j];
      }
    }
    bool pass = true;
    for (i = 0; i < 30; i++) {
      if (quick_s[n - 1][i] != quick_ts[n - 1][i]) {
        pass = false;
      }
    }
    for (i = 1; i < n; i++) {
      for (j = i; j < n; j++) {
        dp[i][j] = INF;
      }
    }
    if (pass)
      cout << solve(n - 1, n - 1) << "\n";
    else
      cout << -1 << "\n";
  }
  return 0;
}
