In the snake exhibition, there are n rooms (numbered 0 to n - 1) arranged in a circle, with a snake in each room. The rooms are connected by n conveyor belts, and the i-th conveyor belt connects the rooms i and (i+1) mod n. In the other words, rooms 0 and 1, 1 and 2, …, n-2 and n-1, n-1 and 0 are connected with conveyor belts.

The i-th conveyor belt is in one of three states:

  * If it is clockwise, snakes can only go from room i to (i+1) mod n. 
  * If it is anticlockwise, snakes can only go from room (i+1) mod n to i. 
  * If it is off, snakes can travel in either direction. 

<image>

Above is an example with 4 rooms, where belts 0 and 3 are off, 1 is clockwise, and 2 is anticlockwise.

Each snake wants to leave its room and come back to it later. A room is returnable if the snake there can leave the room, and later come back to it using the conveyor belts. How many such returnable rooms are there?

Input

Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows. 

The first line of each test case description contains a single integer n (2 ≤ n ≤ 300 000): the number of rooms.

The next line of each test case description contains a string s of length n, consisting of only '<', '>' and '-'.

  * If s_{i} =  '>', the i-th conveyor belt goes clockwise. 
  * If s_{i} =  '<', the i-th conveyor belt goes anticlockwise. 
  * If s_{i} =  '-', the i-th conveyor belt is off. 



It is guaranteed that the sum of n among all test cases does not exceed 300 000.

Output

For each test case, output the number of returnable rooms.

Example

Input


4
4
-&gt;&lt;-
5
&gt;&gt;&gt;&gt;&gt;
3
&lt;--
2
&lt;&gt;


Output


3
5
3
0

Note

In the first test case, all rooms are returnable except room 2. The snake in the room 2 is trapped and cannot exit. This test case corresponds to the picture from the problem statement.

In the second test case, all rooms are returnable by traveling on the series of clockwise belts.
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
import sys
sys.setrecursionlimit(10**4)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
prime=[]
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    # Print the number of two's that divide n 
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2
          
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used 
    for i in range(3,int(math.sqrt(n))+1,2): 
          
        # while i divides n , print i ad divide n 
        while n % i== 0: 
            a.append(i) 
            n = n // i 
              
    # Condition if n is a prime 
    # number greater than 2 
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        s=sin()
        ss=[]
        for i in s:
            if i!="-":
                ss.append(i)
        if len(set(ss))==1:
            print(n)
        else:
            a=[0]*n
            for i in range(n):
                if s[i]=='-':
                    a[i]=1
                    a[(i+1)%n]=1
            print(sum(a))    
        


        
    
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is yferent which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()