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;
int const N = 2010;
int const inf = 1e9;
int dp[N][N], cums[27][N], cumt[27][N];
int n;
int _count(int wh, char ch, int l, int r) {
  if (l > r) return 0;
  r = min(r, n);
  l = min(l, n);
  if (wh == 1) return cums[ch - 'a'][r] - cums[ch - 'a'][l - 1];
  return cumt[ch - 'a'][r] - cumt[ch - 'a'][l - 1];
}
void solve() {
  cin >> n;
  memset(cums, 0, sizeof cums);
  memset(cumt, 0, sizeof cumt);
  string s, t;
  cin >> s >> t;
  map<char, int> cnt;
  for (auto &x : s) cnt[x]++;
  for (auto &x : t) cnt[x]--;
  for (auto x : cnt) {
    if (x.second != 0) {
      cout << "-1\n";
      return;
    }
  }
  s = '#' + s, t = '#' + t;
  for (int ch = 0; ch < 27; ch++) {
    for (int i = 1; i <= n; i++) {
      cums[ch][i] = cums[ch][i - 1] + (s[i] - 'a' == ch);
      cumt[ch][i] = cumt[ch][i - 1] + (t[i] - 'a' == ch);
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (!i) {
        dp[i][j] = 0;
        continue;
      }
      dp[i][j] = inf;
      int &r = dp[i][j];
      r = min(r, dp[i - 1][j] + 1);
      if (s[i] == t[j]) r = min(r, dp[i - 1][j - 1]);
      if (_count(1, t[j], i + 1, n) > _count(2, t[j], j + 1, n)) {
        r = min(r, dp[i][j - 1]);
      }
    }
  }
  cout << dp[n][n] << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}
