Codeforces Round #313 (Div. 2) C. Gerald's Hexagon（補大三角形）

C. 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 ≤ 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.

Output

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

Sample test(s)

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:

題目連接：http://codeforces.com/contest/560/problem/C

題目大意：已知六邊形的六條邊長。求由多少個單位三等邊角形拼成。邊長按順時針順序給出，以單位三角形邊長爲單位。

解題思路：把六邊形補成大等邊三角形。因爲邊長爲n的大三角形由n*n個小單位角形拼成，更easy計算。大三角形減去三個角變成六邊形，減去的三個小三角形的這邊長正好等於六邊形中相應的的三條邊長。所以，設大三角邊長爲s，三個小三角形邊長分別爲a，b，c，那個單位三角形個數爲  s*s-(a*a+b*b+c*c).而 s 正好等於六邊形相鄰三條邊之和，a。b，c 則各自等於六邊形相隔的三條邊，所以很是easy得出答案，可以繪圖加深瞭解。

代碼例如如下：

#include <cstdio>
#include <cstring>
int a[8];
int main()
{
    int s,ans;
    for(int i=0;i<6;i++)
    	scanf("%d",&a[i]);
    s=a[0]+a[1]+a[2];
   	ans=s*s-(a[0]*a[0]+a[2]*a[2]+a[4]*a[4]);
    printf("%d\n",ans);
    return 0;
}