Problem: After his wife's tragic death, Eurydice, Orpheus descended to the realm of death to see her. Reaching its gates was uneasy, but passing through them proved to be even more challenging. Mostly because of Cerberus, the three-headed hound of Hades. 

Orpheus, a famous poet, and musician plans to calm Cerberus with his poetry and safely walk past him. He created a very peculiar poem for Cerberus. It consists only of lowercase English letters. 

We call a poem's substring a palindrome if and only if it reads the same backwards and forwards. A string a is a substring of a string b if a can be obtained from b by deleting several (possibly zero or all) characters from the beginning and several (possibly zero or all) characters from the end.

Unfortunately, Cerberus dislikes palindromes of length greater than 1. For example in the poem abaa the hound of Hades wouldn't like substrings aba and aa.

Orpheus can only calm Cerberus if the hound likes his poetry. That's why he wants to change his poem so that it does not contain any palindrome substrings of length greater than 1.

Orpheus can modify the poem by replacing a letter at any position with any lowercase English letter. He can use this operation arbitrarily many times (possibly zero). Since there can be many palindromes in his poem, he may have to make some corrections. But how many, exactly? Given the poem, determine the minimal number of letters that have to be changed so that the poem does not contain any palindromes of length greater than 1.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 10^5) denoting the number of test cases, then t test cases follow.

The first and only line of each test case contains a non-empty string of lowercase English letters, Orpheus' poem.

The sum of the length of Orpheus' poems in all test cases will not exceed 10^5.

Output

You should output t lines, i-th line should contain a single integer, answer to the i-th test case.

Example

Input


7
babba
abaac
codeforces
zeroorez
abcdcba
bbbbbbb
a


Output


1
1
0
1
1
4
0

Note

In the first test case, we can replace the third character with c and obtain a palindrome-less poem bacba.

In the second test case, we can replace the third character with d and obtain a palindrome-less poem abdac.

In the third test case, the initial poem already doesn't contain any palindromes, so Orpheus doesn't need to change anything there.

Solution (in PYTHON):

Solution:
from __future__ import division, print_function
from itertools import permutations 
import threading,bisect,math,heapq,sys
from collections import deque
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
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()) 
prime=[]
def dfs(n,d,v):
    v[n]=1
    x=d[n]
    for i in x:
        if i not in v:
            dfs(i,d,v)
    return p 
"""**************************MAIN*****************************"""
def main(): 
    t=inin()
    for _ in range(t):
        a=sin()
        a=list(a)
        n=len(a)
        ans=0
        for i in range(1,n):
            if a[i]==a[i-1] or (i>1 and a[i]==a[i-2]):
                ans+=1
                a[i]="*"
        print(ans)
"""***********************************************"""
 
######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
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 = []  
    ll=[]
    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)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
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=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
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
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()