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;
ostream& operator<<(ostream& a, const vector<int>& b) {
  for (auto k : b) cout << k << " ";
  return a;
}
istream& operator>>(istream& a, pair<int, int>& b) {
  cin >> b.first >> b.second;
  return a;
}
ostream& operator<<(ostream& a, const pair<int, int>& b) {
  cout << "{" << b.first << ", " << b.second << "}";
  return a;
}
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
const int INF = 1e9;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
const int MAXN = 2001;
int n;
int dp[MAXN][MAXN];
vector<vector<int>> prefs, preft;
string s, t;
bool have(int i, int j, char k) {
  if (j == 0) return preft[i - 1][k - 'a'];
  return preft[i - 1][k - 'a'] - prefs[j - 1][k - 'a'];
}
void solve() {
  cin >> n;
  cin >> s >> t;
  prefs.assign(n, vector<int>(26, 0));
  preft.assign(n, vector<int>(26, 0));
  prefs[0][s[0] - 'a'] = 1;
  for (int i = 1; i < n; i++) {
    prefs[i] = prefs[i - 1];
    prefs[i][s[i] - 'a']++;
  }
  preft[0][t[0] - 'a'] = 1;
  for (int i = 1; i < n; i++) {
    preft[i] = preft[i - 1];
    preft[i][t[i] - 'a']++;
  }
  if (prefs[n - 1] != preft[n - 1]) {
    cout << "-1\n";
    return;
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      dp[i][j] = INF;
    }
  }
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    dp[i + 1][0] = i + 1;
    for (int j = 0; j <= i; j++) {
      if (have(i + 1, j, s[j]) && dp[i + 1][j + 1] > dp[i + 1][j]) {
        dp[i + 1][j + 1] = dp[i + 1][j];
      }
      if (s[j] == t[i] && dp[i + 1][j + 1] > dp[i][j]) {
        dp[i + 1][j + 1] = dp[i][j];
      }
      if (dp[i + 1][j + 1] > dp[i][j + 1] + 1) {
        dp[i + 1][j + 1] = dp[i][j + 1] + 1;
      }
    }
  }
  cout << dp[n][n] << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    solve();
  }
}
