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(2)
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 2010;
int T, n;
char s[N], t[N];
int dp[N][N];
int sufs[N][30], suft[N][30];
int chs[30], cht[30];
int main() {
  T = read();
  while (T--) {
    n = read();
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) dp[i][j] = 0;
    for (int i = 0; i < 26; i++)
      sufs[n + 1][i] = suft[n + 1][i] = chs[i] = cht[i] = 0;
    for (int i = n; i >= 1; i--) {
      for (int j = 0; j < 26; j++) {
        sufs[i][j] = sufs[i + 1][j];
        suft[i][j] = suft[i + 1][j];
      }
      chs[s[i] - 'a']++;
      cht[t[i] - 'a']++;
      sufs[i][s[i] - 'a']++;
      suft[i][t[i] - 'a']++;
    }
    bool flag = 0;
    for (int j = 0; j < 26; j++) {
      if (chs[j] != cht[j]) flag = 1;
    }
    if (flag) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        dp[i][j] = dp[i][j - 1];
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
        if (s[i] == t[j]) {
          bool flag1 = 0;
          for (int k = 0; k < 26; k++) {
            if (sufs[i + 1][k] < suft[j + 1][k]) {
              flag1 = 1;
            }
          }
          if (!flag1) {
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
          }
        }
      }
    }
    printf("%d\n", n - dp[n][n]);
  }
  return 0;
}
