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 long long int N = 2e3 + 10;
long long int dp[N][N];
long long int suffs[N][26], sufft[N][26];
const long long int inf = 1e10;
long long int n;
string s, t;
long long int dfs(long long int x, long long int y) {
  if (y == 0) {
    return 0;
  }
  long long int &an = dp[x][y];
  if (an != -1) {
    return an;
  }
  an = inf;
  long long int ch = t[y - 1] - 'a';
  if (suffs[x + 1][ch] - sufft[y + 1][ch] > 0) {
    an = min(an, dfs(x, y - 1));
  }
  if (x > 0) {
    an = min(an, dfs(x - 1, y) + 1);
    if (s[x - 1] == t[y - 1]) {
      an = min(an, dfs(x - 1, y - 1));
    }
  }
  return an;
}
void solve() {
  cin >> n;
  cin >> s >> t;
  for (long long int i = 0; i <= n; i++) {
    for (long long int j = 0; j < 26; j++) {
      suffs[i][j] = sufft[i][j] = 0;
    }
  }
  for (long long int i = 0; i <= n; i++) {
    for (long long int j = 0; j <= n; j++) {
      dp[i][j] = -1;
    }
  }
  for (long long int i = n; i >= 1; i--) {
    for (long long int j = 0; j < 26; j++) {
      sufft[i][j] = sufft[i + 1][j];
      suffs[i][j] = suffs[i + 1][j];
    }
    suffs[i][s[i - 1] - 'a']++;
    sufft[i][t[i - 1] - 'a']++;
  }
  long long int ans = dfs(n, n);
  if (ans >= inf) {
    cout << -1 << "\n";
  } else {
    cout << ans << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
