Polycarp came up with a new programming language. There are only two types of statements in it: 

  * "x := s": assign the variable named x the value s (where s is a string). For example, the statement var := hello assigns the variable named var the value hello. Note that s is the value of a string, not the name of a variable. Between the variable name, the := operator and the string contains exactly one space each.
  * "x = a + b": assign the variable named x the concatenation of values of two variables a and b. For example, if the program consists of three statements a := hello, b := world, c = a + b, then the variable c will contain the string helloworld. It is guaranteed that the program is correct and the variables a and b were previously defined. There is exactly one space between the variable names and the = and + operators. 



All variable names and strings only consist of lowercase letters of the English alphabet and do not exceed 5 characters.

The result of the program is the number of occurrences of string haha in the string that was written to the variable in the last statement.

Polycarp was very tired while inventing that language. He asks you to implement it. Your task is — for given program statements calculate the number of occurrences of string haha in the last assigned variable.

Input

The first line contains an integer t (1 ≤ t ≤ 10^3). Then t test cases follow.

The first line of each test case contains a single integer n (1 ≤ n ≤ 50) — the number of statements in the program. All variable names and strings are guaranteed to consist only of lowercase letters of the English alphabet and do not exceed 5 characters.

This is followed by n lines describing the statements in the format described above. It is guaranteed that the program is correct.

Output

For each set of input data, output the number of occurrences of the haha substring in the string that was written to the variable in the last statement.

Example

Input


4
6
a := h
b := aha
c = a + b
c = c + c
e = c + c
d = a + c
15
x := haha
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
x = x + x
1
haha := hah
5
haahh := aaaha
ahhhh = haahh + haahh
haahh = haahh + haahh
ahhhh = ahhhh + haahh
ahhaa = haahh + ahhhh


Output


3
32767
0
0

Note

In the first test case the resulting value of d is hhahahaha.
import time

#start_time = time.time()
#def TIME_(): print(time.time()-start_time)

import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string, heapq as h
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        import os
        self.os = os
        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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
            self.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")

def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def getBin(): return list(map(int,list(input())))
def isInt(s): return '0' <= s[0] <= '9'
def ceil_(a,b): return (a+b-1)//b

MOD = 10**9 + 7

"""
For each variable I need to know how many hahas it currently has, what its first three letters are, and what its last three letters are
"""

def solve():
    global case
    case += 1
    memo = dd(int)
    memo2 = dd(str)
    memo3 = dd(str)
    N = getInt()
    for i in range(N):
        S = getStrs()
        #if case == 66:
         #   print(' '.join(S))
        if len(S) == 3:
            memo[S[0]] = S[2].count('haha')
            memo2[S[0]] = S[2][:3]
            memo3[S[0]] = S[2][-3:]
        else:
            memo[S[0]] = memo[S[2]] + memo[S[4]]
            tmp = memo3[S[2]] + memo2[S[4]]
            for i in range(len(tmp)):
                if tmp[i:i+4] == 'haha':
                    memo[S[0]] += 1
            L2, L4 = len(memo2[S[2]]), len(memo3[S[4]])
            memo2[S[0]] = memo2[S[2]] + (memo2[S[4]][:(3 - L2)] if 3 > L2 else '')
            memo3[S[0]] = (memo3[S[2]][-(3 - L4):] if 3 > L4 else '') + memo3[S[4]]
        lst = S[0]
        #print(memo,memo2,memo3,sep='\n')
    return memo[lst]
#ahhahahaah+ahaah
global case
case = 0

for _ in range(getInt()):
    if case <= 5 or case >= 65:
        print(solve())
    else:
        print(solve())

#TIME_()