10300

German farmers are given a premium depending on the conditions at their farmyard. Imagine the following simplified regulation: you know the size of each farmer's farmyard in square meters and the number of animals living at it. We won't make a difference between different animals, although this is far from reality. Moreover you have information about the degree the farmer uses environment-friendly equipment and practices, expressed in a single integer greater than zero. The amount of money a farmer receives can be calculated from these parameters as follows. First you need the space a single animal occupies at an average. This value (in square meters) is then multiplied by the parameter that stands for the farmer's environment-friendliness, resulting in the premium a farmer is paid per animal he owns. To compute the final premium of a farmer just multiply this premium per animal with the number of animals the farmer owns.

Input

The first line of input contains a single positive integer n (<20), the number of test cases. Each test case starts with a line containing a single integer f (0<f<20), the number of farmers in the test case. This line is followed by one line per farmer containing three positive integers each: the size of the farmyard in square meters, the number of animals he owns and the integer value that expresses the farmer’s environment-friendliness. Input is terminated by end of file. No integer in the input is greater than 100000 or less than 0.

Output

For each test case output one line containing a single integer that holds the summed burden for Germany's budget, which will always be a whole number. Do not output any blank lines.

Sample Input

3

5

1 1 1

2 2 2

3 3 3

2 3 4

8 9 2

3

9 1 8

6 12 1

8 1 1

3

10 30 40

9 8 5

100 1000 70

Sample Output

38

86

7445

题意：总的保险费=一个动物占用面积×环境友好度×动物数量=农场面积×环境友好度；输入的三个值为：农场面积，动物数量，环境友好度。

思路：其实将第一的乘第三个再加起来就可以。

#include<stdio.h>
int main()
{
int n,m,i,j,a,b,c,max;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&m);
max=0;
for(j=0;j<m;j++)
{
scanf("%d%d%d",&a,&b,&c);
int d;
d=a*c;
max+=d;
}
printf("%d\n",max);
}
return 0;
}