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 inf = 2e9;
const long long INF = 8e18;
const int maxn = 2e3 + 5;
int n;
string s, t;
int dp[maxn][maxn], suf[maxn][26], suf2[maxn][26];
bool vis[maxn][maxn];
inline int dfs(int i, int j) {
  if (vis[i][j]) return dp[i][j];
  vis[i][j] = 1;
  if (s[i - 1] == t[j - 1]) dp[i][j] = min(dp[i][j], dfs(i - 1, j - 1));
  if (suf[i + 1][t[j - 1] - 'a'] > suf2[j + 1][t[j - 1] - 'a'])
    dp[i][j] = min(dp[i][j], 1 + dfs(i, j - 1));
  if (i > 0) dp[i][j] = min(dp[i][j], dfs(i - 1, j));
  return dp[i][j];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> s >> t;
    for (int i = 0; i < n + 5; ++i)
      for (int j = 0; j < n + 5; ++j) vis[i][j] = 0, dp[i][j] = inf;
    for (int i = 0; i < n + 5; ++i)
      for (int j = 0; j < 26; ++j) suf[i][j] = suf2[i][j] = 0;
    for (int i = n; i >= 1; --i) {
      for (int j = 0; j < 26; ++j) {
        suf[i][j] = suf[i + 1][j];
        suf2[i][j] = suf2[i + 1][j];
      }
      ++suf[i][s[i - 1] - 'a'];
      ++suf2[i][t[i - 1] - 'a'];
    }
    vis[0][0] = 1, dp[0][0] = 0;
    dfs(n, n);
    if (dp[n][n] == inf)
      cout << -1 << '\n';
    else
      cout << dp[n][n] << '\n';
  }
  return 0;
}
