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.

The first and the single line of the input contains 6 space-separated integers a₁, a₂, a₃, a₄, a₅ and a₆ (1 ≤ a_i ≤ 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.

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

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：按顺序给出一个各内角均为120°的六边形的六条边长，求该六边形能分解成多少个边长为1的单位三角形。

题解：把六边形补全为一个三角行，然后求出这个三角形可以分成的边长为1的等边三角形的个数再减去刚刚补上的那部分（也就是那三个小三角形）

这里的关键就是等边三角形由小三角形组成的个数是n*n，这里的n是边长......

代码如下：

 #include <stdio.h>

 int main()

 {

     int b1,b2,b3,b4,b5,b6;

     scanf("%d%d%d%d%d%d",&b1,&b2,&b3,&b4,&b5,&b6);

     int ans=(b2+b3+b4)*(b2+b3+b4)-(b2*b2+b4*b4+b6*b6);

     printf("%d\n",ans);

     return ;

 }