CodeForces 559A Gerald's Hexagon

A. Gerald's Hexagon

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to 

.
Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.

He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his counting.
Help the boy count the triangles.

Input

The first and the single line of the input contains 6 space-separated integers a1, a2, a3, a4, a5 and a6 (1 ≤ ai ≤ 1000)
— the lengths of the sides of the hexagons in centimeters in the clockwise order. It is guaranteed that the hexagon with the indicated properties and the exactly such sides exists.

Output

Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split.

Examples

input

1 1 1 1 1 1

output

6

input

1 2 1 2 1 2

output

13

Note

This is what Gerald's hexagon looks like in the first sample:



And that's what it looks like in the second sample:



题意：给你一个各角都是120度的六边形，问这个六边形可以划分为多少个边长为1的等边三角行

思路：六边形的面积除以三角形的面积。



等角六边形的对边平行，连接两个对边的定点，就得到一个梯形和两个三角形，梯形的上底是a[4]，下底是a[1]，

高由划红线的地方算出a[3]*sin(PI/3.0)+a[2]*sin(PI/3.0)，这样就可以算出梯形的面积，其余的两个三角形的

面积为a[2]*a[3]*sin(PI/3.0)/2， a[5]*a[6]*sin(PI/3.0)/2小三角形的面积：1*1*sin(PI/3.0)/2.0
;

#include<stdio.h>
#include<math.h>
int main()
{
double a[10],sum;
int i;
while(scanf("%lf%lf%lf%lf%lf%lf",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=EOF)
{
sum=(a[1]+a[4])*(a[3]+a[2])+a[2]*a[3]+a[5]*a[6];
printf("%.lf\n",sum);

}
return 0;
}