B. Opposites Attract

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Everybody knows that opposites attract. That is the key principle of the "Perfect Matching" dating agency. The "Perfect Matching" matchmakers have classified each registered customer by his interests and assigned to the i-th client number t_i( - 10 ≤ t_i ≤ 10). Of course, one number can be assigned to any number of customers.

"Perfect Matching" wants to advertise its services and publish the number of opposite couples, that is, the couples who have opposite values of t. Each couple consists of exactly two clients. The customer can be included in a couple an arbitrary number of times. Help the agency and write the program that will find the sought number by the given sequence t₁, t₂, ..., t_n. For example, if t = (1,  - 1, 1,  - 1), then any two elements t_i and t_j form a couple if i and j have different parity. Consequently, in this case the sought number equals 4.

Of course, a client can't form a couple with him/herself.

Input

The first line of the input data contains an integer n (1 ≤ n ≤ 10⁵) which represents the number of registered clients of the "Couple Matching". The second line contains a sequence of integers t₁, t₂, ..., t_n ( - 10 ≤ t_i ≤ 10), t_i — is the parameter of the i-th customer that has been assigned to the customer by the result of the analysis of his interests.

Output

Print the number of couples of customs with opposite t. The opposite number for x is number  - x (0 is opposite to itself). Couples that only differ in the clients' order are considered the same.

Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.

Sample test(s)

input

5
-3 3 0 0 3

output

3

input

3
0 0 0

output

3

Note

In the first sample the couples of opposite clients are: (1,2), (1,5) и (3,4).

In the second sample any couple of clients is opposite.

解题说明：此题就是统计一组数中有多少对数加起来和为0，最简单的方法是用两层循环，但是此题中n的数目很大，很容易超时。认真读题，发现ti的值限于[-10,10]之间，所以不应该记录每一个ti的值，而是统计ti中取每个值的个数有多少，然后就可以通过两个互为相反数的出现次数计算出和为0的次数。不过由于题目中存在负数，而数组下标肯定不能为负，统一给下标加上一个10即可。边输入边统计。

b[0]  -10出现的次数

b[1]  -9出现的次数

...

b[10] 0出现的次数

...

b[19] 9出现的次数

b[20] 10出现的次数

最后就是挨个统计b[i]*b[20-i]的值，注意到b[10]中的0是自己配对，也就是排列组合中的Cn^2，单独计算即可。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;

int main()
{
	int n,i,j;
	int a;
	double b[21];
	double sum;
	sum=0;
	scanf("%d",&n);
	for(i=0;i<21;i++)
	{
		b[i]=0;
	}
	for(i=0;i<n;i++)
	{
		scanf("%d",&a);
		b[a+10]++;
	}
	for(i=0;i<10;i++)
	{
		sum+=b[i]*b[20-i];
	}
	sum+=b[10]*(b[10]-1)/2;

	printf("%.0lf\n",sum);
	
	return 0;
}