A. Sea Battle

In order to make the "Sea Battle" game more interesting, Boris decided to add a new ship type to it. The ship consists of two rectangles. The first rectangle has a width of

The rectangles are placed on field in the following way:

the second rectangle is on top the first rectangle;
they are aligned to the left, i.e. their left sides are on the same line;
the rectangles are adjacent to each other without a gap.

See the pictures in the notes: the first rectangle is colored red, the second rectangle is colored blue.

Formally, let's introduce a coordinate system. Then, the leftmost bottom cell of the first rectangle has coordinates

After the ship is completely destroyed, all cells neighboring by side or a corner with the ship are marked. Of course, only cells, which don't belong to the ship are marked. On the pictures in the notes such cells are colored green.

Find out how many cells should be marked after the ship is destroyed. The field of the game is infinite in any direction.

Input

Four lines contain integers

Output

Print exactly one integer — the number of cells, which should be marked after the ship is destroyed.

Examples

input

2 1 2 1

output

input

2 2 1 2

output

Note

In the first example the field looks as follows (the first rectangle is red, the second rectangle is blue, green shows the marked squares):

$\text{[math]}$

In the second example the field looks as:

$\text{[math]}$

题意：输入4个数，前两个数表示第一个矩阵的长和宽，后两个数表示第二个矩阵的长和宽。然后找出这两个矩阵旁边的相邻单位的数量。就像上图一样，红色代表第一个矩阵，蓝色代表第二个，首先这两个矩阵必须时相邻的，然后找出旁边绿色方块的数量。我的思路就是先算出第一个矩阵的所以相邻个数，再算出第二个矩阵的相邻个数，然后再减去它们两个共同所占据的个数。

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

int main()
{
    int w1,w2,h1,h2;
    cin>>w1>>h1>>w2>>h2;
    int sum1=w1*2+h1*2+4;    //第一个矩阵相邻单位的数量，加4的意思是4个角
    int sum2=w2*2+h2*2+4;        //第二个矩阵相邻单位的数量
    int sum=sum1+sum2;                //两个矩阵的和
    sum-=min(w1,w2)*2+4;                //减去他们共同所占据的相邻单位，找两个矩阵相邻的最小边，那里才是它们的共同占据的数量单位，这里加4的意思是减去两个矩阵分别多出的两个角
    cout<<sum<<endl;
    return 0;
}