Problem:
Score : 400 points

Problem Statement

Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.

He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:

a < b + c

b < c + a

c < a + b

How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Constraints

All values in input are integers.

3 \leq N \leq 2 \times 10^3

1 \leq L_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 ... L_N

Constraints

Print the number of different triangles that can be formed.

Sample Input 1

4
3 4 2 1

Sample Output 1

1

Only one triangle can be formed: the triangle formed by the first, second, and third sticks.

Sample Input 2

3
1 1000 1

Sample Output 2

0

No triangles can be formed.

Sample Input 3

7
218 786 704 233 645 728 389

Sample Output 3

23

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define debug_interval ','
#define dump_interval ' '
#define debug_toggle 1
//{
#include<bits/stdc++.h>
using namespace std;
#define hi cerr<<"hi"<<endl;
#define int long long
#define INT_MAX LLONG_MAX
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define fi first
#define se second
#define mp make_pair
#define rev reverse
#define dans dump(ans)
#define MOD 1000000007
#define amp(v,n) (v).count(n)?v[n]++:v[n]=1
#define sysp system("pause")
#define PI acos(-1)
#define pf push_front
#define ins insert

//{
inline int toInt(string s){int v;istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x){ostringstream sout;sout<<x;return sout.str();}
template<class...A> inline void dump(){cout<<endl;}
template<class...A> inline void dump_rest() {cout<<endl;}
template<class T, class...A> inline void dump_rest(const T& first, const A&... rest){cout<<dump_interval<<first;dump_rest(rest...);}
template<class T,class...A> inline void dump(const T&first,const A&...rest){cout<<first;dump_rest(rest...);}
template<class...A> inline void debug(){cerr<<endl;}
template<class...A> inline void debug_rest() {cerr<<endl;}
template<class T, class...A> inline void debug_rest(const T& first, const A&... rest){cerr<<debug_interval<<first;debug_rest(rest...);}
template<class T,class...A> inline void debug(const T&first,const A&...rest){if(debug_toggle)cerr<<first,debug_rest(rest...);}
unsigned XorShift(void){static unsigned x = 123456789;static unsigned y = 362436069;static unsigned z = 521288629;static unsigned w = 88675123;unsigned t;t = x ^ (x << 11);x = y; y = z; z = w;return w = (w ^ (w >> 19)) ^ (t ^ (t >> 8));}
//}

typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}
template<int m>class mint{private:int i;public:mint() : i(0){}mint(int i): i((i%m+m)%m){}mint operator+(const mint& o){return o.i+i;}mint operator*(const mint& o){return o.i*i;}mint operator-(){return -i;}operator int() {return i;}};

//}

main(){
	int n,ans=0;cin>>n>>ans;
	int l[n];rep(i,n)cin>>l[i];sort(l,l+n);
	rep(i,n)rep(j,i){
		ans+=lower_bound(l,l+n,l[i]+l[j])-l-i-1;
	}
	dans;
}