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 Python (3.4.3) (judge verdict: Accepted)

import bisect

n = int(input())
l = list(map(int, input().split()))
l.sort()

ans = 0

for i in range(n):
    for j in range(i + 1, n):
        bise = bisect.bisect_left(l, l[i] + l[j])
        if bise > j:
            ans += bise - (j + 1)
        else:
            pass

print(ans)


# import bisect
#
# N = int(input())
# Ls = list(map(int, input().split()))
# Ls.sort()
#
# # O(N^2 * logN)の全探索で解く
# ans = 0
# for a_i in range(N):
#     for b_i in range(a_i+1, N):
#         c_i = bisect.bisect_left(Ls, Ls[a_i]+Ls[b_i]) # 左側に境界がある右側のインデックスが帰ってくる
#         if c_i > b_i:
#             ans += c_i - b_i - 1
#         else:
#             pass
#
# print(ans)

# from itertools import combinations
# import numpy as np
#
# n = int(input())
# hoge = list(map(int, input().split()))
# hoge.sort()
#
# # hoge = np.sort(l)
# ll = []
# ans = 0
#
# def triangle(target, v1, v2):
#     return target < v1 + v2
#
# for i in range(len(hoge)):
#     for j in range(i + 1, len(hoge)):
#         left = hoge[j]
#         right = hoge[-1] + 1
#         while left + 1 < right:
#             mid = (left + right) / 2
#             if triangle(mid, hoge[i], hoge[j]):
#                 ans += len(hoge[hoge >= mid])
#                 break
#             else:
#                 left = mid
#
# print(ans)

"""3回目
# 事前の各要素xより大きい要素個数を求める
for v in hoge:
    ll.append([v, len(hoge[v <= hoge])])

# 事前にコンビネーションを確保する。ループはしない！
combi = np.array([[sum(v), v[1]] for v in combinations(hoge, 2)])

print(combi)
print(combi[:, 0])
print(ll)

# 要素のループをもう一度回し、各
for v, cnt in ll:
    print(v, cnt, len(combi[(v < combi[:, 0]) & (combi[] < v)]), sep=',')
    ans += cnt * len(combi[v < combi])

print(ans)

"""


"""2回目 comnbinationループがだめ
for v in combinations(hoge, 2):
    s = sum(v)
    ans += len(hoge[(v[1] < hoge) & (hoge < s)])

"""



"""1
これだとTLEが出る
combinationの全パターンを数え上げると遅すぎる
ソート済み配列 × なにかで組み合わせを数え上げる必要

from itertools import combinations

n = int(input())
l = map(int, input().split())

ans = 0

for v in combinations(l, 3): # combination を使ってループした時点ででn^2
    if v[0] < v[1] + v[2] and v[1] < v[0] + v[2] and v[2] < v[0] + v[1]:
        ans += 1

print(ans)
"""