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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

const (
	initialBufSize = 1e4
	maxBufSize     = 1e6 + 7
)

var buf []byte = make([]byte, maxBufSize)
var sc = bufio.NewScanner(os.Stdin)

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(buf, maxBufSize)
}

func main() {
	N := nextInt()
	Ln := make([]int, N)
	for i := range Ln {
		Ln[i] = nextInt()
	}
	var ans int
	for i := 0; i < N; i++ {
		for j := i + 1; j < N; j++ {
			for k := j + 1; k < N; k++ {
				if Ln[i] < Ln[j]+Ln[k] && Ln[j] < Ln[i]+Ln[k] && Ln[k] < Ln[i]+Ln[j] {
					ans++
				}
			}
		}
	}
	fmt.Println(ans)
}

func nextLine() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func abs(a int) int {
	if a > 0 {
		return a
	} else {
		return -a
	}
}

func pow(p, q int) int {
	res := p
	if q == 0 {
		return 1
	}
	for i := 0; i < q-1; i++ {
		res *= p
	}
	return res
}

func MinOf(vars ...int) int {
	min := vars[0]
	for _, i := range vars {
		if min > i {
			min = i
		}
	}
	return min
}

func MaxOf(vars ...int) int {
	max := vars[0]
	for _, i := range vars {
		if max < i {
			max = i
		}
	}
	return max
}

func gcdof2numbers(a int, b int) int {
	if b == 0 {
		return a
	}
	return gcdof2numbers(b, a%b)
}

func lcmof2numbers(a int, b int) int {
	return a / gcdof2numbers(a, b) * b
}

// マイナスの場合は考慮していない
func fuctorial(a int) int {
	if a == 1 || a == 0 {
		return 1
	} else {
		return fuctorial(a-1) * a
	}
}

func NextPermutation(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}

// 素因数分解したmapを返す
func primeFuctorize(n int) map[int]int {
	pf := map[int]int{}
	for i := 2; i*i <= n; i++ {
		for n%i == 0 {
			pf[i]++
			n /= i
		}
	}
	if n != 1 {
		pf[n]++
	}
	return pf
}