# Candies and Two Sisters

There are two sisters Alice and Betty. You have $n$ candies. You want to distribute these $n$ candies between two sisters in such a way that:

• Alice will get $a$ ($a > 0$) candies;
• Betty will get $b$ ($b > 0$) candies;
• each sister will get some integer number of candies;
• Alice will get a greater amount of candies than Betty (i.e. $a > b$);
• all the candies will be given to one of two sisters (i.e. $a+b=n$).

Your task is to calculate the number of ways to distribute exactly $n$ candies between sisters in a way described above. Candies are indistinguishable.

Formally, find the number of ways to represent $n$ as the sum of $n=a+b$, where $a$ and $b$ are positive integers and $a>b$.

You have to answer $t$ independent test cases.Input

The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases. Then $t$ test cases follow.

The only line of a test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^9$) — the number of candies you have.Output

For each test case, print the answer — the number of ways to distribute exactly $n$ candies between two sisters in a way described in the problem statement. If there is no way to satisfy all the conditions, print $0$.Exampleinput

6
7
1
2
3
2000000000
763243547


output

3
0
0
1
999999999
381621773


Note

For the test case of the example, the $3$ possible ways to distribute candies are:

• $a=6$, $b=1$;
• $a=5$, $b=2$;
• $a=4$, $b=3$.

Solution:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
cout << (n - 1) / 2 << '\n';
}
return 0;
}