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 maxn = 2000 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int dp[maxn][maxn];
int cnts[26][maxn], cntt[26][maxn];
char s[maxn], t[maxn];
int solve(int i, int j) {
  if (i == 0 || j == 0) return 0;
  int &ans = dp[i][j];
  if (ans != -1) return ans;
  ans = 1e9;
  if (i > 0) {
    ans = 1 + solve(i - 1, j);
    if (s[i] == t[j]) {
      ans = min(ans, solve(i - 1, j - 1));
    }
  }
  int ch = t[j] - 'a';
  if (cnts[ch][i + 1] - cntt[ch][j + 1] > 0) ans = min(ans, solve(i, j - 1));
  return ans;
}
int main() {
  int cs;
  scanf("%d", &cs);
out:
  while (cs--) {
    int n;
    scanf("%d", &n);
    scanf("%s%s", s + 1, t + 1);
    for (int i = 1; i <= n + 1; ++i) {
      for (int j = 0; j < 26; ++j) cnts[j][i] = cntt[j][i] = 0;
      for (int j = 1; j <= n + 1; ++j) dp[i][j] = -1;
    }
    for (int i = n; i >= 1; --i) {
      for (int j = 0; j < 26; ++j) {
        cnts[j][i] = cnts[j][i + 1];
        cntt[j][i] = cntt[j][i + 1];
      }
      cnts[s[i] - 'a'][i]++;
      cntt[t[i] - 'a'][i]++;
    }
    for (int i = 0; i < 26; ++i) {
      if (cnts[i][1] != cntt[i][1]) {
        printf("-1\n");
        goto out;
      }
    }
    printf("%d\n", solve(n, n));
  }
  return 0;
}
