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;
const int inf = 1e8;
const int maxn = 2e3 + 10;
string s, t;
int p[26][maxn], q[26][maxn];
int mem[maxn][maxn];
int n;
int dp(int x, int y) {
  if (y == n) return 0;
  if (x > n) return inf;
  int &ans = mem[x][y];
  if (ans != -1) return ans;
  ans = 1 + dp(x + 1, y);
  if (s[x] == t[y]) {
    ans = min(ans, dp(x + 1, y + 1));
  }
  int c = t[y] - 'a';
  if (p[c][x] > q[c][y]) ans = min(ans, dp(x, y + 1));
  return ans;
}
void solve() {
  cin >> n;
  cin >> s >> t;
  reverse(s.begin(), s.end());
  reverse(t.begin(), t.end());
  int cntP[26], cntQ[26];
  memset(cntP, 0, sizeof cntP);
  memset(cntQ, 0, sizeof cntQ);
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 26; j++) {
      p[j][i] = cntP[j];
      q[j][i] = cntQ[j];
    }
    if (i == n) continue;
    cntP[s[i] - 'a'] += 1;
    cntQ[t[i] - 'a'] += 1;
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      mem[i][j] = -1;
    }
  }
  int ans = dp(0, 0);
  if (ans >= inf) ans = -1;
  cout << ans << '\n';
}
int main(int argc, char const *argv[]) {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
