Codeforces Round #313 (Div. 2) C. Geralds Hexagon

Description

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 Input

Input

1 1 1 1 1 1

Output

6

Input

1 2 1 2 1 2

Output

13

Hint

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

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

题目大意：已知六边形的六条边长，求由多少个单位三等边角形拼成。边长按顺时针顺序给出，以单位三角形边长为单位

把六边形补成大等边三角形

求三角形比较容易

在用大三角形减去3个补上去的小三角形

#include<iostream>
using namespace std;
int main(){
    int x[6];
    while(cin>>x[0]>>x[1]>>x[2]>>x[3]>>x[4]>>x[5]){
     cout<<(x[0]+x[2]+x[1])*(x[0]+x[2]+x[1])-x[0]*x[0]-x[2]*x[2]-x[4]*x[4]<<endl;
    }
return  0;
}