C. Lengthening Sticks
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given three sticks with positive integer lengths of a, b, and c centimeters. You can increase length of some of them by some positive integer number of centimeters (different sticks can be increased by a different length), but in total by at most l centimeters. In particular, it is allowed not to increase the length of any stick.

Determine the number of ways to increase the lengths of some sticks so that you can form from them a non-degenerate (that is, having a positive area) triangle. Two ways are considered different, if the length of some stick is increased by different number of centimeters in them.

Input

The single line contains 4 integers a, b, c, l (1 ≤ a, b, c ≤ 3·105, 0 ≤ l ≤ 3·105).

Output

Print a single integer — the number of ways to increase the sizes of the sticks by the total of at most l centimeters, so that you can make a non-degenerate triangle from it.

Examples
Input 
1 1 1 2
Output 
4
Input 
1 2 3 1
Output 
2
Input 
10 2 1 7
Output 
0
Note

In the first sample test you can either not increase any stick or increase any two sticks by 1 centimeter.

In the second sample test you can increase either the first or the second stick by one centimeter. Note that the triangle made from the initial sticks is degenerate and thus, doesn't meet the conditions.

【题意】给出a,b,c,L，要求a+x,b+y,c+z构成三角形，x+y+z<=L，问有多少中分法(x,y,z可为0)。

【分析】不会这个题，题解可以看这个大牛的文章，http://blog.csdn.net/u013486414/article/details/47950919，我完全没想到这方法，总感觉是搜索什么的，实在太弱了。看了这题解才觉得这真的好题啊，考了一手脑洞和思维。解题方法参见上面blog。

【AC 代码】

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL solve(LL a,LL b,LL c,LL l)
{
    LL maxx = a+b+c+l;
    LL ans = 0;
    for(LL i=a; i<=a+l; i++){
        if(b+c>i) continue;
        else{
            LL x = min(i,maxx-i)-(b+c);
            ans += (x+2)*(x+1)/2;
        }
    }
    return ans;
}
int main()
{
    LL a,b,c,l;
    LL ans;
    scanf("%lld%lld%lld%lld",&a,&b,&c,&l);
    ans = (l+1)*(l+2)*(l+3)/6;
    ans -= solve(a,b,c,l);
    ans -= solve(b,a,c,l);
    ans -= solve(c,a,b,l);
    printf("%lld\n",ans);
    return 0;
}