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 N = 2.2e3;
int n, cnt[N], dp[N][N], sa[N][26], sb[N][26];
char a[N], b[N];
int DP(int x, int y) {
  if (y > n) return 0;
  if (x > y) return N * N;
  for (int i = 0; i < 26; i++)
    if (sa[x][i] < sb[y][i]) return N * N;
  if (dp[x][y] != -1) return dp[x][y];
  if (a[x] == b[y]) return dp[x][y] = DP(x + 1, y + 1);
  return dp[x][y] = min(DP(x, y + 1) + 1, DP(x + 1, y));
}
int mian() {
  memset(cnt, 0, sizeof(cnt));
  memset(sa, 0, sizeof(sa)), memset(sb, 0, sizeof(sb));
  cin >> n;
  scanf("%s", a + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = -1;
  for (int i = 1; i <= n; i++) {
    cnt[a[i]]++;
  }
  scanf("%s", b + 1);
  for (int i = 1; i <= n; i++) cnt[b[i]]--;
  for (int i = 'a'; i <= 'z'; i++) {
    if (cnt[i]) return puts("-1"), 0;
  }
  int now = n + 1, ans = 0;
  for (int i = n; i; i--) {
    for (int j = 0; j < 26; j++)
      sa[i][j] = sa[i + 1][j], sb[i][j] = sb[i + 1][j];
    sa[i][a[i] - 'a']++;
    sb[i][b[i] - 'a']++;
  }
  cout << DP(1, 1) << endl;
  return 0;
}
int main() {
  int T;
  cin >> T;
  while (T--) mian();
}
