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 = 2005;
char s[N], t[N];
int dp[N][N];
int num[256];
bool isAnagram(int n) {
  memset(num, 0, sizeof(num));
  for (int i = 1; i <= n; ++i) {
    ++num[s[i]];
    --num[t[i]];
  }
  for (int i = 0; i < 256; ++i) {
    if (num[i] != 0) {
      return false;
    }
  }
  return true;
}
int pre[2][N][26];
int solveDp(int x, int y) {
  if (y == 0) {
    return 0;
  }
  int &ret = dp[x][y];
  if (ret != -1) {
    return ret;
  }
  ret = 1e9;
  if (x) {
    ret = 1 + solveDp(x - 1, y);
    if (s[x] == t[y]) {
      ret = min(ret, solveDp(x - 1, y - 1));
    }
  }
  int al = t[y] - 'a';
  if (pre[0][x + 1][al] > pre[1][y + 1][al]) {
    ret = min(ret, solveDp(x, y - 1));
  }
  return ret;
}
void solve2() {
  int n;
  scanf("%d", &n);
  scanf(" %s %s ", s + 1, t + 1);
  if (!isAnagram(n)) {
    printf("-1\n");
    return;
  }
  for (int i = 0; i < 26; ++i) {
    pre[0][n + 1][i] = pre[1][n + 1][i] = 0;
  }
  for (int i = n; i; --i) {
    for (int j = 0; j < 26; ++j) {
      pre[0][i][j] = pre[0][i + 1][j];
    }
    ++pre[0][i][s[i] - 'a'];
  }
  for (int i = n; i; --i) {
    for (int j = 0; j < 26; ++j) {
      pre[1][i][j] = pre[1][i + 1][j];
    }
    ++pre[1][i][t[i] - 'a'];
  }
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) {
      dp[i][j] = -1;
    }
  }
  printf("%d\n", solveDp(n, n));
}
void solve() {
  int n;
  scanf("%d", &n);
  while (n--) {
    solve2();
  }
}
int main() {
  solve();
  return 0;
}
