C - Maximum GCD

Let’s consider all integers in the range from 1 to n (inclusive).

Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a,b), where 1≤a<b≤n.

The greatest common divisor, gcd(a,b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.

Input
The first line contains a single integer t (1≤t≤100) — the number of test cases. The description of the test cases follows.

The only line of each test case contains a single integer n (2≤n≤106).

Output
For each test case, output the maximum value of gcd(a,b) among all 1≤a<b≤n.

Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1,2)=gcd(2,3)=gcd(1,3)=1.

In the second test case, 2 is the maximum possible value, corresponding to gcd(2,4).

#include<stdio.h>
#include<stdlib.h>
#include <math.h>
int main()
{ int m,n;
scanf("%d",&m);
while(~scanf("%d",&n))
{
    printf("%d\n",n/2);
}

    return 0;
}