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 = 2e3 + 10;
int dp[maxn][maxn];
char s[maxn], t[maxn];
int cnt[2][maxn][26];
int dfs(int x, int y) {
  int &now = dp[x][y];
  if (now != -1) return now;
  if (y == 0) return now = 0;
  now = INT_MAX;
  if (x > 0) {
    if (s[x] == t[y]) now = min(now, dfs(x - 1, y - 1));
    now = min(now, dfs(x - 1, y) + 1);
  }
  if (cnt[0][x + 1][t[y] - 'a'] > cnt[1][y + 1][t[y] - 'a']) {
    now = min(now, dfs(x, y - 1));
  }
  return now;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d%s%s", &n, s + 1, t + 1);
    memset(cnt, 0, sizeof cnt);
    int flag = 0;
    for (int i = n; i >= 1; --i) {
      memcpy(cnt[0][i], cnt[0][i + 1], sizeof cnt[0][i]);
      memcpy(cnt[1][i], cnt[1][i + 1], sizeof cnt[1][i]);
      ++cnt[0][i][s[i] - 'a'];
      ++cnt[1][i][t[i] - 'a'];
    }
    for (int i = 0; i < 26; ++i)
      if (cnt[0][1][i] != cnt[1][1][i]) flag |= 1;
    if (flag)
      puts("-1");
    else {
      for (int i = 1; i <= n; ++i) {
        for (int j = i; j <= n; ++j) dp[i][j] = -1;
      }
      printf("%d\n", dfs(n, n));
    }
  }
  return 0;
}
