链接：https://codeforces.com/contest/1263/problem/A

You have three piles of candies: red, green and blue candies:

  • the first pile contains only red candies and there are rr candies in it,
  • the second pile contains only green candies and there are gg candies in it,
  • the third pile contains only blue candies and there are bb candies in it.

Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can't eat two candies of the same color in a day.

Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.

Input

The first line contains integer tt (1≤t≤10001≤t≤1000) — the number of test cases in the input. Then tt test cases follow.

Each test case is given as a separate line of the input. It contains three integers rr, gg and bb (1≤r,g,b≤1081≤r,g,b≤108) — the number of red, green and blue candies, respectively.

Output

Print tt integers: the ii-th printed integer is the answer on the ii-th test case in the input.

Example

input

Copy

6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8

output

Copy

1
2
2
10
5
9

Note

In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.

In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.

In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.

题解：这是俺第三次碰到这个类型的题了，第一次是comet国庆欢乐赛俺被卡了几乎整场，还好室友提点了我一句；第二次是沈阳打铁；这是第三次，前两次都磕磕绊绊，所以印象深刻。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<map>
#include<set>
#include<algorithm>
#include<cstdlib>
using namespace std;
long long n,b,t,g,r,l,k,y,x,s=0;
long long a[3000001];
int main()
{
   	cin>>t;
   	while(t--)
	{
		cin>>a[1]>>a[2]>>a[3];
		sort(a+1,a+1+3);
		if(a[3]>a[1]+a[2])
		cout<<a[1]+a[2]<<endl;
		else
		cout<<(a[1]+a[2]+a[3])/2<<endl;
	}
    return 0;
}