Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)A. Friends Meeting

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
找到中间数用求和公式算一下就行了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define pn printf("
")
using namespace std;
typedef long long ll;

const int maxn = 1e5+7;

int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    int mid = abs(a-b);
    int aa = mid/2 + mid%2, bb = mid/2;
    printf("%I64d
", 1LL*(1+aa)*aa/2 + 1LL*(1+bb)*bb/2);
}