F(1)=1 F(2)=3 F(N)=F(N-1)-F(N-2) Now you are given N, you have to find the value of F(1)+F(2)+......+F(N) .


Input

Input starts with an integer T(1<=T<=1000), denoting the number of test cases. Each test case contains an integer N(1<=N<=10^18).

Output

For each test case, print the value.

Example


Input: 2 1 2 Output: 1 4



思路：


尝试写出几个答案你就会发现，f[n]是一个循环的数列，以6为周期，每个周期内的六个数相加为0，这样只要对n进行取6的模就可以输出答案。 



#include<iostream>
#include<cstdio>
using namespace std;
int a[7];
int main(){
  a[1] = 1;a[2] = 4;a[3] = 6;
  a[4] = 5; a[5] = 2; a[0] = 0;
  int T;
  long long n,te;
  scanf("%d",&T);
  while(T--){
    scanf("%lld",&n);
    te = n % 6;
    cout<<a[te]<<endl; 
  }
  return 0; 
}