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;
constexpr int MAXN = 2e3 + 3, INF = 0x3f3f3f3f;
int tc, n, sufs[MAXN][26], suft[MAXN][26], dp[MAXN][MAXN], vis[MAXN][MAXN];
string s, t;
int DP(int i, int j) {
  if (i == 0 || j == 0) return 0;
  if (j < i) return INF;
  if (vis[i][j]) return dp[i][j];
  vis[i][j] = 1;
  dp[i][j] = DP(i - 1, j) + 1;
  if (s[i] == t[j]) dp[i][j] = min(dp[i][j], DP(i - 1, j - 1));
  if (sufs[i + 1][t[j] - 'a'] > suft[j + 1][t[j] - 'a'])
    dp[i][j] = min(dp[i][j], DP(i, j - 1));
  return dp[i][j];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n >> s >> t;
    s = " " + s, t = " " + t;
    memset(sufs, 0, sizeof(sufs));
    memset(suft, 0, sizeof(suft));
    for (int i = n; i; i--)
      for (int j = 0; j < 26; j++)
        sufs[i][j] = sufs[i + 1][j] + (s[i] - 'a' == j),
        suft[i][j] = suft[i + 1][j] + (t[i] - 'a' == j);
    bool tag = true;
    for (int i = 0; i < 26; i++)
      if (sufs[1][i] != suft[1][i]) tag = false;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) dp[i][j] = -1, vis[i][j] = 0;
    tag ? cout << DP(n, n) << "\n" : cout << "-1\n";
  }
  return 0;
}
