CF1217A Creating a Character

You play your favourite game yet another time. You chose the character you didn't play before. It has str points of strength and int points of intelligence. Also, at start, the character has exp free experience points you can invest either in strength or in intelligence (by investing one point you can either raise strength by 1 or raise intelligence by 1).

Since you'd like to make some fun you want to create a jock character, so it has more strength than intelligence points (resulting strength is strictly greater than the resulting intelligence).

Calculate the number of different character builds you can create (for the purpose of replayability) if you must invest all free points. Two character builds are different if their strength and/or intellect are different.

Input

The first line contains the single integer T (1≤T≤100) — the number of queries. Next T lines contain descriptions of queries — one per line.

This line contains three integers str, int and exp (1≤str,int≤10^8,0≤exp≤10^8) — the initial strength and intelligence of the character and the number of free points, respectively.

Output

Print TT integers — one per query. For each query print the number of different character builds you can create.

Example

input

4
5 3 4
2 1 0
3 5 5
4 10 6

output

3
1
2
0

Note

In the first query there are only three appropriate character builds: (str=7,int=5), (8,4) and (9,3). All other builds are either too smart or don't use all free points.

In the second query there is only one possible build: (2,1).

In the third query there are two appropriate builds: (7,6), (8,5).

In the fourth query all builds have too much brains.

先解释下题意：给定T行输入，每行输入三个整数str，int，exp。每点exp可以增加一点str或int，求str>int的方案共有多少种。

显然我们只要求出最少要加几点str使str>int就可以求出答案了。

下面给出代码，这里是用二分来求解。

#include <bits/stdc++.h>
using namespace std;
long long a,b,c;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
        cin>>a>>b>>c;
        int l=0,r=c;
        while(l<=r)
        {
            int mid=(l+r)/2;
            if(a+mid>b+c-mid)
            {
                r=mid-1;
            }
            else
            {
                l=mid+1;
            }
        }
        cout<<c-l+1<<endl;
    }
    return 0;
}