#include <iostream>

using namespace std;

int main()
{
	int x, y;
	cout << "x = ";
	cin >> x;
	cout << "y = ";
	cin >> y;
	
	int n, v = max(abs(x), abs(y)), g = 8 * (v * (v + 1) / 2);

	if (y == v || x == v) { // Top row or right column
		n = g - x + y - 6 * v;
	}
	else { // Left column or bottom row
		n = g + x - y - 2 * v;
	}

	printf("n = %d", n);

	return 0;
}

/* --------------
 * Task condition
 * --------------
 * You have a spiral of whole numbers, starting from 0, like this:
 *
 * 36 < 35 < 34 < 33 < 32 < 31 < 30
 * V                              ^
 * 37   16 < 15 < 14 < 13 < 12   29
 * V     V                   ^    ^
 * 38   17    4 <  3 <  2   11   28
 * V     V    V         ^    ^    ^
 * 39   18    5    0 >  1   10   27  ...
 * V     V    V              ^    ^    ^
 * 40   19    6 >  7 >  8 >  9   26   51
 * V     V                        ^    ^
 * 41   20 > 21 > 22 > 23 > 24 > 25   50
 * V                                   ^
 * 42 > 43 > 44 > 45 > 46 > 47 > 48 > 49
 *
 * The distance between two neighbouring numbers is 1, meaning 0 is at (0;0), 1 is at (1;0), 38 is at (-3;1).
 * You are given the coordinates of a number and you have to find the actual number.
 *
 * --------------------------
 * Explanation of my solution
 * --------------------------
 *
 * My solution is a very mathematical one. First, I divide the spiral into squares,
 * depending on the offset of 0;0.
 * Example:
 * - Square 1 (because x and y are between -1 and 1 (one of them is always 1 or -1)):
 *  4 <  3 <  2
 *  V         ^
 *  5         1
 *  V          
 *  6 >  7 >  8
 *
 * - Square 2 (because x and y are between -2 and 2 (one of them is always 2 or -2)):
 * 16 < 15 < 14 < 13 < 12
 *  V                   ^
 * 17                  11
 *  V                   ^
 * 18                  10
 *  V                   ^
 * 19                   9
 *  V                    
 * 20 > 21 > 22 > 23 > 24
 *
 * - Square 3 (because x and y are between -3 and 3 (one of them is always 3 or -3)):
 * 36 < 35 < 34 < 33 < 32 < 31 < 30
 * V                              ^
 * 37                            29
 * V                              ^
 * 38                            28
 * V                              ^
 * 39                            27
 * V                              ^
 * 40                            26
 * V                              ^
 * 41                            25
 * V                               
 * 42 > 43 > 44 > 45 > 46 > 47 > 48
 *
 * And so on. I call the number of the square "v".
 * We can find it by getting the absolute value of both coordinates and get the biggest number.
 *
 * Then there is the biggest number in the square, the one in the bottom right corner, I call that "g".
 * g for square 1 (v = 1) is 8, for v = 2: g = 24, for v = 3: g = 48 and so on.
 * We can see that each g is divisble by 8, but the multiplier is slightly tricky to find.
 * The multiplier is the sum of all whole number from 1 up until v.
 *
 * Example:
 * v = 2
 * g = 8 * (Sum of numbers from 1 to v) = 8 * (1 + 2) = 8 * 3 = 24
 * 
 * v = 3
 * g = 8 * (Sum of numbers from 1 to v) = 8 * (1 + 2 + 3) = 8 * 6 = 48
	  * 
	  * Then, we divide the whole square into two corners. We take the diagonal from top left to bottom right, and check in which part the coordinates are.
	  * We do that because we have two formulas. In brief, what they do is tranform the coordinates into the number from which we'll subtract from g.
	  *
	  * Example:
	  * v = 3, g = 48
	  *
	  * Lets take a look at the formula for the bottom of the square. We aren't interested in g, so the important part is "x - y - 2*v".
	  * What it does is make the coordinates for x and y correspond to a number from -12 to 0.
	  * So this:
	  * 36
	  * V
	  * 37
	  * V
	  * 38
	  * V
	  * 39
	  * V
	  * 40
	  * V
	  * 41
	  * V
	  * 42 > 43 > 44 > 45 > 46 > 47 > 48
	  * Becomes:
	  * -12
	  *  V
	  * -11
	  *  V
	  * -10
	  *  V
	  * -9
	  *  V
	  * -8
	  *  V
	  * -7
	  *  V
	  * -6 > -5 > -4 > -3 > -2 > -1 > 0
	  *
	  * "- 2 * v" is -6, the number in the bottom left corner. Then, if we increase Y (meaning we want to go up the left column),
	  * we want the number to also decrease (make it further away from 0). We also have to account for X, because there it's always -3.
	  * So, we need them to sum up to 0 when we're on the bottom left corner, so we can just subtract Y from X. As we said,
	  * X is always -3 on the left column, so by increasing Y, we also increase the negative number.
	  *
	  * Ok, here are some example numbers:
	  * X  | Y  | X - Y
	  * ---|----|------
	  * -3 | -3 |  0
	  * -3 | -2 | -1
	  * -3 |  0 | -3
	  * 
	  * This actually works out for us just fine when we're traversing the bottom row. There, Y is always -3, and
	  * when increasing X we want to increase our number (make it close to 0). So, X - Y become a positive number,
	  * which summed up with the negative number ("-2*v") makes the result number closer to 0, which is what we want.
	  *
	  * Here are more example numbers:
	  * X  | Y  | X - Y
	  * ---|----|------
	  * -3 | -3 | 0
	  * -2 | -3 | 1
	  *  0 | -3 | 3
	  *
	  * The formula for the top of the square (top row and right column) works the same , but there our corner number isn't "-2*v" but "-6*v",
	  * and we want the reverse effect: increase number (make smaller) on increasing X and decerease number on incrasing Y,
	  * so we make "X - Y" to be "Y - X"
	  *
	  * -13 < -14 < -15 < -16 < -17 < -18 < -19
	  *                                      ^
	  *                                     -20
	  *                                      ^
	  *                                     -21
	  *                                      ^
	  *                                     -22
	  *                                      ^
	  *                                     -23
	  *                                      ^
	  *                                     -24
 */