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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long N = 2005, inf = 1e15;
long long dp[N][N], suffix1[N][26], suffix2[N][26];
long long n;
string s, t;
long long dpsolve(long long i, long long j) {
  if (j == 0) {
    return 0;
  }
  if (dp[i][j] != -1) return dp[i][j];
  long long ans = inf;
  if (i > 0) {
    ans = min(ans, 1 + dpsolve(i - 1, j));
    if (s[i] == t[j]) ans = min(ans, dpsolve(i - 1, j - 1));
  }
  if (suffix1[i + 1][t[j] - 'a'] >= suffix2[j][t[j] - 'a'])
    ans = min(ans, dpsolve(i, j - 1));
  dp[i][j] = ans;
  return ans;
}
void solve() {
  cin >> n;
  cin >> s;
  cin >> t;
  s = ' ' + s;
  t = ' ' + t;
  for (long long i = n; i >= 1; i--) {
    for (long long j = 0; j < 26; j++) {
      suffix1[n + 1][j] = suffix2[n + 1][j] = 0;
      suffix1[i][j] = (s[i] - 'a' == j) + suffix1[i + 1][j];
      suffix2[i][j] = (t[i] - 'a' == j) + suffix2[i + 1][j];
    }
  }
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= n; j++) dp[i][j] = -1;
  bool flag = true;
  for (long long j = 0; j < 26; j++)
    if (suffix1[1][j] != suffix2[1][j]) flag = false;
  if (!flag)
    cout << "-1\n";
  else {
    long long ans = dpsolve(n, n);
    cout << ans << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
