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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int N = 2e3 + 5;
const long long INF = 1e18;
const long double EPS = 10e-9;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long random(long long a, long long b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
long long suffix[N][26][2];
long long dp[N][N];
bool done[N][N];
string s, t;
long long DP(long long i, long long j) {
  if (i == -1) return 0;
  if (done[i][j]) return dp[i][j];
  if (i >= 0) {
    dp[i][j] = 1 + DP(i - 1, j);
    if (s[i] == t[j]) dp[i][j] = min(dp[i][j], DP(i - 1, j - 1));
  }
  int cur = t[j] - 'a';
  if (suffix[i + 1][cur][0] - suffix[j + 1][cur][1] > 0)
    dp[i][j] = min(dp[i][j], DP(i, j - 1));
  done[i][j] = true;
  return dp[i][j];
}
void solve() {
  long long n;
  cin >> n;
  cin >> s >> t;
  long long freq[2][26];
  memset(freq, 0, sizeof(freq));
  for (long long i = 0; i < n; i++) {
    freq[0][s[i] - 'a']++;
    freq[1][t[i] - 'a']++;
  }
  for (long long i = 0; i < 26; i++) {
    if (freq[0][i] == freq[1][i]) continue;
    cout << -1 << endl;
    return;
  }
  for (long long i = 0; i < 26; i++)
    for (long long j = 0; j < 2; j++) suffix[n][i][j] = 0;
  for (long long i = n - 1; i >= 0; i--) {
    for (long long j = 0; j < 26; j++)
      for (long long k = 0; k < 2; k++) suffix[i][j][k] = suffix[i + 1][j][k];
    suffix[i][s[i] - 'a'][0]++;
    suffix[i][t[i] - 'a'][1]++;
  }
  for (long long i = 0; i < n + 1; i++)
    for (long long j = 0; j < n + 1; j++) {
      dp[i][j] = INF;
      done[i][j] = false;
    }
  long long ans = DP(n - 1, n - 1);
  if (ans >= INF) cout << -1 << endl;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
