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: Accepted)


#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#include <prettyprint.hpp>
#define debug(...)  cerr << "[" << #__VA_ARGS__ << "]: ", d_err(__VA_ARGS__);
#else
#define debug(...) 83;
#endif

void d_err() {
    cerr << endl;
}

template <typename H, typename... T>
void d_err(H h, T... t) {
    cerr << h << " ";
    d_err(t...);
}

template <typename T>
void print(T x) {
    cout << x << endl;
}

#define ALL(x) (x).begin(), (x).end()
#define FOR(i, m, n) for (int i = (m); i < (n); ++i)
#define REVFOR(i, m, n) for (int i = (n - 1); i >= (m); --i)
#define REP(i, n) FOR(i, 0, n)
#define REVREP(i, n) REVFOR(i, 0, n)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define bcnt __builtin_popcountll

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<ll,ll> Pll;
typedef pair<int,int> Pin;

ll INF = 1e16;
int inf = 1e9;
ll MOD = 1e9+7;

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    
    int N;
    cin >> N;

    vll L(N);
    REP(i, N) cin >> L[i];

    vector<int> C(1e4, 0);
    REP(i, N) C[L[i]]++;
    FOR(i, 1, C.size()) C[i] += C[i - 1];

    ll ans = 0;
    REP(i, N) FOR(j, i+1, N) {
        ll a = L[i], b = L[j];

        ll l = max(a - b, b - a);
        ll r = a + b - 1;

        if (C[r] - C[l] < 1) continue;

        ans += C[r] - C[l];

        if (l < a && a <= r) ans--;
        if (l < b && b <= r) ans--;
    }

    print(ans / 3);

    
}