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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll        = long long;
using ld        = long double;
using vl        = vector<ll>;
using vd        = vector<ld>;
using vs        = vector<string>;
using vb        = vector<bool>;
using vvl       = vector<vector<ll>>;
using vvd       = vector<vector<ld>>;
using vvs       = vector<vector<string>>;
using vvb       = vector<vector<bool>>;
using pll       = pair<ll, ll>;
constexpr ll mod= 1e9 + 7;
#define ALL(x) (x).begin(), (x).end()
#define REP(i, n) for(ll(i)= 0; (i) < (n); (i)++)
#define REPS(i, n) for(ll(i)= 1; (i) <= (n); (i)++)
#define RREP(i, n) for(ll(i)= (n - 1); (i) >= 0; (i)--)
#define RREPS(i, n) for(ll(i)= (n); (i) > 0; (i)--)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define UNIQUE(v) v.erase(unique(ALL(v)), v.end())
#define YES(n) ((n) ? "YES" : "NO")
#define Yes(n) ((n) ? "Yes" : "No")
#define yes(n) ((n) ? "yes" : "no")

template <class T>
inline void chmin(T& a, T b) {
    if(a > b) { a= b; }
}

template <class T>
inline void chmax(T& a, T b) {
    if(a < b) { a= b; }
}

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(12);
    ll N;
    cin >> N;
    vl L(N);
    REP(i,N)cin >> L[i];
    sort(ALL(L));
    ll ans = 0;
    REP(i,N){
        for(int l = i+1;l < N;l++){
            ans += N - (lower_bound(ALL(L),L[l]) - L.begin() + 1) - (L.end() - lower_bound(ALL(L),L[i] + L[l]));
        }
    }
    cout << ans << "\n";
}