'''

Created on May 6, 2015

@author: Chocolate

'''

import abc

'''

When taking the following rules into account:

even + even = even

even + odd = odd

odd + even = odd

odd + odd = even

The parity of the first Fibonacci numbers is:

o o e o o e o o e ...

Thus basically, you simply need to do steps of three. Which is:

(1,1,2)

(3,5,8)

(13,21,34)

Given (a,b,c) this is (b+c,b+2*c,2*b+3*c).

This means we only need to store the two last numbers, and calculate given (a,b), (a+2*b,2*a+3*b).

Thus (1,2) -> (5,8) -> (21,34) -> ... and always return the last one.

This will work faster than a "filter"-approach because that uses the if-statement which reduces pipelining.

def checkTheMultiple(arr, num, n):

# print("In checker");

i = 2;

a = num * i;

while a <= n:

# print(a);

arr[a - 1] = 1;

# print(arr);

i = i + 1;

a = num * i;

def checkPrime(n):

resultlist=[];

if n >= 2:

llist = [0] * n;

for j in xrange(1, n, 1):

if llist[j] == 0:

checkTheMultiple(llist, j + 1, n);

# print("Back");

# print(llist);

resultlist.extend([j+1]);

return resultlist;

'''

def fibo(n):

a=[0]*(n+1);

a[0]=1;

a[1]=2;

for i in xrange(2,n+1,2):

a[i]=a[i-2]+(a[i-1]*2);

a[i+1]=(2*a[i-2])+(a[i-1]*3);

if a[i+1]>=n:

break;

return a;

for _ in xrange(input()):

N=input();

abc=[x for x in fibo(N) if x % 2 == 0];

#print fibo(N);

bca=[x for x in abc if x<N];

#print abc;

print(sum(bca));