Codeforces Round 56-A. Dice Rolling(思维题）

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mishka got a six-faced dice. It has integer numbers from 22 to 77 written on its faces (all numbers on faces are different, so this is an almostusual dice).

Mishka wants to get exactly xx points by rolling his dice. The number of points is just a sum of numbers written at the topmost face of the dice for all the rolls Mishka makes.

Mishka doesn't really care about the number of rolls, so he just wants to know any number of rolls he can make to be able to get exactly xxpoints for them. Mishka is very lucky, so if the probability to get xx points with chosen number of rolls is non-zero, he will be able to roll the dice in such a way. Your task is to print this number. It is guaranteed that at least one answer exists.

Mishka is also very curious about different number of points to score so you have to answer tt independent queries.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of queries.

Each of the next tt lines contains one integer each. The ii-th line contains one integer xixi (2≤xi≤1002≤xi≤100) — the number of points Mishka wants to get.

Output

Print tt lines. In the ii-th line print the answer to the ii-th query (i.e. any number of rolls Mishka can make to be able to get exactly xixi points for them). It is guaranteed that at least one answer exists.

Example

input

Copy

4
2
13
37
100

output

Copy

1
3
8
27

Note

In the first query Mishka can roll a dice once and get 22 points.

In the second query Mishka can roll a dice 33 times and get points 55, 55 and 33 (for example).

In the third query Mishka can roll a dice 88 times and get 55 points 77 times and 22 points with the remaining roll.

In the fourth query Mishka can roll a dice 2727 times and get 22 points 1111 times, 33 points 66 times and 66 points 1010 times.

每次都是两道，唉，何时才能提高啊

题解：把每次都看成2即可，感觉有点水

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;


int main()
{
	
	int n;
	cin>>n;
	int k;
	for(int t=0;t<n;t++)
    {
    	cin>>k;
    	cout<<k/2<<endl;
	 } 
	
	return 0;
}