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 comment(linker, "/STACK:1024000000, 1024000000")
#pragma GCC optimize(3)
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-10;
const int MAXN = 2e3 + 5;
const long long MOD = 998244353LL;
char s[2][MAXN];
int dp[MAXN][MAXN];
int suf[2][MAXN][26];
int main() {
  int T;
  scanf("%d", &T);
  for (int cs = 1; cs <= T; ++cs) {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < 2; ++i) {
      scanf("%s", s[i] + 1);
      for (int j = 0; j < 26; ++j) {
        suf[i][n + 1][j] = 0;
      }
      for (int j = n; j >= 1; --j) {
        for (int k = 0; k < 26; ++k) {
          suf[i][j][k] = suf[i][j + 1][k];
        }
        ++suf[i][j][s[i][j] - 'a'];
      }
    }
    for (int i = 0; i <= n; ++i) {
      for (int j = 0; j <= n; ++j) {
        dp[i][j] = INF;
      }
    }
    for (int i = 0; i <= n; ++i) {
      for (int j = i; j <= n; ++j) {
        if (j == 0) {
          dp[i][j] = 0;
        } else {
          if (i > 0) {
            if (s[0][i] == s[1][j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
            dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
          }
          int ch = s[1][j] - 'a';
          if (suf[0][i + 1][ch] > suf[1][j + 1][ch])
            dp[i][j] = min(dp[i][j], dp[i][j - 1]);
        }
      }
    }
    int ans = dp[n][n];
    if (ans == INF) ans = -1;
    printf("%d\n", ans);
  }
  return 0;
}
