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 N = 2003 + 7, M = 1000000007, inf = 1e9;
const long long INF = LLONG_MAX;
int n, Q;
int suf[26][N][2], dp[N][N];
string s, t;
int getans(int i, int j) {
  assert(i <= j);
  if (j == 0) {
    assert(i == 0);
    return 0;
  }
  int &mafia = dp[i][j];
  if (mafia != -1) return mafia;
  int a1 = inf, a2 = inf, a3 = inf;
  if (i) {
    a1 = 1 + getans(i - 1, j);
    if (s[i] == t[j]) a2 = getans(i - 1, j - 1);
  }
  int x = t[j] - 'a';
  if (suf[x][i + 1][0] > suf[x][j + 1][1])
    a3 = getans(i, j - 1);
  else
    ;
  return mafia = min(a1, min(a2, a3));
}
void solve() {
  cin >> n >> s >> t;
  s = " " + s;
  t = " " + t;
  for (int i = n; i >= 1; i--) {
    int x = s[i] - 'a';
    int y = t[i] - 'a';
    for (int ch = 0; ch < (int)26; ch++) {
      suf[ch][i][0] = (i < n ? suf[ch][i + 1][0] : 0) + (x == ch);
      suf[ch][i][1] = (i < n ? suf[ch][i + 1][1] : 0) + (y == ch);
    }
  }
  for (int i = 0; i < (int)n + 1; i++)
    for (int j = 0; j < (int)n + 1; j++) dp[i][j] = -1;
  int ans = getans(n, n);
  if (ans > n) ans = -1;
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  cin >> test;
  while (test--) solve();
}
