count-x
Source: hackerearth
Difficulty: EASY

Each digit of a number is represented by a cluster of adjacent X, with spaces between neighboring clusters. The digit is determined by the number of X in the cluster. For example, the number 243 would be represented by the following pattern
     -XX-XXXX-XXX-
So to represent 0 the pattern will be -- (2 dashes without any X in between).
Given a string as input  you have to output the corresponding number.
NOTE - All the number will fit into 32 bit integer. And string does not contain 10 consecutive 'X's.

First line contains number of test cases.
The first line of every test case contains one string.

Constraints-
1 ≤ t ≤ 30
3 ≤ string length ≤ 50

SAMPLE INPUT
2
-XX-XXXX-XXX-
-XX--XXXX---XXX-

SAMPLE OUTPUT
243
204003


t=eval(input())
for j in range(0,t):
	string=input().split("-")
	num=0
	for k in range(1,len(string)-1):
		num=num*10+len(string[k])
	print(num)


t = int(input())
for _ in range(t):
	lst = []
	s = input()
	lst = s.split("-")
	counter = []
	for i in lst:
		count = 0
		for j in i:
			count+=1
		counter.append(count)
	a = ''.join(map(str,counter))
	b = ''
	for i in range(1,len(a)-1):
		b = b+a[i]
	print(b)
		
			
			


t=eval(input(""))
while t>0 and t<31:
	c=0
	k=[]
	s=input("")
	l=len(s)
	for i in range(0,l,1):
		if s[i]=='X':
			c=c+1
			if s[i+1]=='-':
				k.append(c)
		elif s[i]=='-':
			c=0
			if i+1<l: 
				if s[i+1]=='-':
					k.append(c)
	print(''.join(map(str,k)))
	t=t-1
		
	


t = eval(input())
o=[]
for i in range(t):
	l = input()
	num=''
	j=0
	while j < len(l):
		if l[j]=='-':
			j+=1
			while j<len(l) and l[j]=='-':
				num+='0'
				j+=1
		else:
			c=0
			while j<len(l) and l[j]=='X':
				c+=1
				j+=1
			num+=str(c)

	o.append(int(num))

for e in o:
	print(e)


t = eval(input())
while t > 0:
	number = input()
	temp = number.split('-')
	orig = []
	for value in range(1, len(temp) - 1):
		orig.append(str(len(temp[value])))
	print((''.join(orig)))
	t -= 1