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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T sqr(T a) {
  return a * a;
}
template <class T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
const double pi = acos(-1.0);
int n;
int m;
string s, t;
const int inf = 1e9;
vector<vector<int> > to, cnts, cntt, f;
void init() {
  f = vector<vector<int> >(n + 1, vector<int>(n + 1, -1));
  to = vector<vector<int> >(n + 1, vector<int>(128, -1));
  cnts = vector<vector<int> >(n + 1, vector<int>(128, 0));
  cntt = vector<vector<int> >(n + 1, vector<int>(128, 0));
  for (int i = n - 1; i >= 0; i--) {
    for (int c = 'a'; c <= 'z'; c++) {
      to[i][c] = to[i + 1][c];
    }
    to[i][t[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    for (int c = 'a'; c <= 'z'; c++) {
      cnts[i + 1][c] = cnts[i][c];
      cntt[i + 1][c] = cntt[i][c];
    }
    cnts[i + 1][s[i]]++;
    cntt[i + 1][t[i]]++;
  }
}
int go(int i, int j) {
  if (i == n && j == n) return 0;
  if (i >= n) return inf;
  if (f[i][j] != -1) return f[i][j];
  int cnt1 = 0, cnt2 = 0;
  cnt1 = cntt[j][s[i]];
  cnt2 = cnts[i][s[i]];
  int res = inf;
  if (cnt1 > cnt2) res = min(res, go(i + 1, j) + 1);
  int k = to[j][s[i]];
  if (k != -1) res = min(res, go(i + 1, k + 1));
  return f[i][j] = res;
}
int main() {
  int nt;
  cin >> nt;
  for (int tt = 1; tt <= nt; tt++) {
    cin >> n >> s >> t;
    init();
    int ans = go(0, 0);
    if (ans < inf) {
      cout << ans << endl;
    } else {
      cout << -1 << endl;
    }
  }
  return 0;
}
