题干：

Two friends are on the coordinate axis Ox in points with integer coordinates. One of them is in the point x1 = a, another one is in the point x2 = b.

Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1 + 2 + 3 = 6.

The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.

Input

The first line contains a single integer a (1 ≤ a ≤ 1000) — the initial position of the first friend.

The second line contains a single integer b (1 ≤ b ≤ 1000) — the initial position of the second friend.

It is guaranteed that a ≠ b.

Output

Print the minimum possible total tiredness if the friends meet in the same point.

Examples

Input

3
4

Output

1

Input

101
99

Output

2

Input

5
10

Output

9

Note

In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.

In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.

In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9.

题目大意：

     数轴上有两人分别在a,b两点，每个人每走一个单位坐标，增加疲劳值，且递增，如左边的人走三步，疲劳值为1+2+3=6。两个人都能走，现希望两人碰面，且求两个人的最小疲劳值的和。

解题报告：

  其实很好想啦，如果距离是偶数，找中间点就是了。如果是奇数，那就先找中间点-1，然后单独算一个人多走的一步。

这里的距离可以用他们的坐标的奇偶来看。是一样的。

 

AC代码：

#include<bits/stdc++.h>using namespace std;
long long ans;int main()
{int a,b;cin>>a>>b;int sum = a+b;int half;if(sum%2 == 0) {half = sum/2;sum = abs(half-a);for(int i = 1; i<=sum; i++) {ans +=i;}ans+=ans;}else {int i;half = sum/2;a=min(a,b);sum = abs(half-a);for(i = 1; i<=sum; i++) {ans+=i;}ans+=ans;ans+=i;}printf("%lld\n",ans);return 0 ;
}