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.

題意：一個篩子寫著2----7六個數字，問轉多少次轉的和可以準確的得到X。（次數任意）

PS：這個題週三晚上的訓練比賽上出現了。當時也不知道這是個簽到難度的題，自己思維也想錯了，所以才寫這篇部落格的

分析：其他的不用考慮直接n/2輸出就可以了，當時自己多想了對7取模特判的，

程式碼：

#include<iostream>
using namespace std;
int main()
{

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

	return 0;
}