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;
long long n;
string s, t;
const long long MX = 2005;
bool used[MX][MX];
long long dp[MX][MX];
long long acumS[MX][26];
long long acumT[MX][26];
const long long INF = 1e18;
long long getAcumS(long long l, long long r, char c) {
  if (l > r) return 0;
  return acumS[r + 1][c - 'a'] - acumS[l][c - 'a'];
}
long long getAcumT(long long l, long long r, char c) {
  if (l > r) return 0;
  return acumT[r + 1][c - 'a'] - acumT[l][c - 'a'];
}
long long minMove(long long i, long long j) {
  if (i > j) return INF;
  if (i == -1) return 0;
  if (used[i][j]) return dp[i][j];
  used[i][j] = true;
  long long ans = INF;
  if (s[i] == t[j]) {
    ans = min(ans, minMove(i - 1, j - 1));
  }
  if (getAcumS(i + 1, n - 1, t[j]) > getAcumT(j + 1, n - 1, t[j])) {
    ans = min(ans, minMove(i, j - 1));
  }
  ans = min(ans, minMove(i - 1, j) + 1);
  return dp[i][j] = ans;
}
void clear(long long n) {
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      used[i][j] = false;
    }
  }
}
void solve() {
  cin >> n;
  clear(n);
  cin >> s >> t;
  for (long long i = 0; i < n; i++) {
    for (long long let = 0; let < 26; let++) {
      acumS[i + 1][let] = acumS[i][let];
      acumT[i + 1][let] = acumT[i][let];
    }
    acumS[i + 1][s[i] - 'a']++;
    acumT[i + 1][t[i] - 'a']++;
  }
  string a, b;
  a = s;
  b = t;
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  if (a != b) {
    cout << "-1\n";
    return;
  }
  cout << minMove(n - 1, n - 1) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T;
  cin >> T;
  for (long long i = 0; i < (long long)T; i++) solve();
  return 0;
}
