human
You are given a keyboard that consists of 26 keys. The keys are arranged sequentially in one row in a certain order. Each key corresponds to a unique lowercase Latin letter.

You have to type the word s on this keyboard. It also consists only of lowercase Latin letters.

To type a word, you need to type all its letters consecutively one by one. To type each letter you must position your hand exactly over the corresponding key and press it.

Moving the hand between the keys takes time which is equal to the absolute value of the difference between positions of these keys (the keys are numbered from left to right). No time is spent on pressing the keys and on placing your hand over the first letter of the word.

For example, consider a keyboard where the letters from 'a' to 'z' are arranged in consecutive alphabetical order. The letters 'h', 'e', 'l' and 'o' then are on the positions 8, 5, 12 and 15, respectively. Therefore, it will take |5 - 8| + |12 - 5| + |12 - 12| + |15 - 12| = 13 units of time to type the word "hello". 

Determine how long it will take to print the word s.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases.

The first line of a description contains a keyboard — a string of length 26, which consists only of lowercase Latin letters. Each of the letters from 'a' to 'z' appears exactly once on the keyboard.

The second line of the description contains the word s. The word has a length from 1 to 50 letters inclusive and consists of lowercase Latin letters.

Output

Print t lines, each line containing the answer to the corresponding test case. The answer to the test case is the minimal time it takes to type the word s on the given keyboard.

Example

Input

5
abcdefghijklmnopqrstuvwxyz
hello
abcdefghijklmnopqrstuvwxyz
i
abcdefghijklmnopqrstuvwxyz
codeforces
qwertyuiopasdfghjklzxcvbnm
qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq
qwertyuiopasdfghjklzxcvbnm
abacaba

Output

13
0
68
0
74
from 
Can you provide a Python program to determine how long it will take to print the word s?
assistant
#Here is Python program to determine how long it will take to print the word s.
math import *
from collections import *
import os
from io import BytesIO, IOBase
import sys
 
sys.setrecursionlimit(10 ** 6)
MOD = 1000000007
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
# Code by Big Dick Daddy Dick

def binpow(a, b, m):
    a %= m
    x = 1
    while b > 0:
        if b & 1:
            x = x * a % m
        a = a * a % m
        b >>= 1
    return x

def sakurajima(n):
    d = defaultdict(list)
   ...