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.
import sys

int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]

def main():
    inf=10**9
    for _ in range(II()):
        n=II()
        s = SI()
        t = SI()

        cnts=[[0]*26 for _ in range(n+1)]
        cntt=[[0]*26 for _ in range(n+1)]
        for i in range(n-1,-1,-1):
            cnts[i][ord(s[i])-97]+=1
            cntt[i][ord(t[i])-97]+=1
            for j in range(26):
                cnts[i][j]+=cnts[i+1][j]
                cntt[i][j]+=cntt[i+1][j]

        ng=False
        for j in range(26):
            if cnts[0][j]!=cntt[0][j]:
                ng=True
                break
        if ng:
            print(-1)
            continue

        dp=[[inf]*(n+1) for _ in range(n+1)]
        for i in range(n+1):dp[i][0]=0
        for i in range(n+1):
            for j in range(i,n+1):
                if i==n and j==n:break
                pre=dp[i][j]
                if pre==inf:continue
                if i<j and i+1<=n:dp[i+1][j]=min(dp[i+1][j],pre+1)
                if i+1<=n and j+1<=n and s[i]==t[j]:dp[i+1][j+1]=min(dp[i+1][j+1],pre)
                if j+1<=n and cnts[i+1][ord(t[j])-97]>cntt[j+1][ord(t[j])-97]:
                    dp[i][j + 1] = min(dp[i][j + 1], pre)
        #p2D(dp)
        print(dp[n][n])

main()
