#include <algorithm>

#include <cstring>

#include <string.h>

#include <iostream>

#include <list>

#include <map>

#include <set>

#include <stack>

#include <string>

#include <utility>

#include <vector>

#include <cstdio>

#include <cmath>

#define LL long long

#define N 200005

#define INF 0x3ffffff

using namespace std;

const double PI = acos(-1.0);

struct Complex // 复数

{

    double r,i;

    Complex(double _r = ,double _i = )

    {

        r = _r; i = _i;

    }

    Complex operator +(const Complex &b)

    {

        return Complex(r+b.r,i+b.i);

    }

    Complex operator -(const Complex &b)

    {

        return Complex(r-b.r,i-b.i);

    }

    Complex operator *(const Complex &b)

    {

        return Complex(r*b.r-i*b.i,r*b.i+i*b.r);

    }

};

void change(Complex y[],int len) // 二进制平摊反转置换 O(logn)

{

    int i,j,k;

    for(i = , j = len/;i < len-;i++)

    {

        if(i < j)swap(y[i],y[j]);

        k = len/;

        while( j >= k)

        {

            j -= k;

            k /= ;

        }

        if(j < k)j += k;

    }

}

void fft(Complex y[],int len,int on) //DFT和FFT

{

    change(y,len);

    for(int h = ;h <= len;h <<= )

    {

        Complex wn(cos(-on**PI/h),sin(-on**PI/h));

        for(int j = ;j < len;j += h)

        {

            Complex w(,);

            for(int k = j;k < j+h/;k++)

            {

                Complex u = y[k];

                Complex t = w*y[k+h/];

                y[k] = u+t;

                y[k+h/] = u-t;

                w = w*wn;

            }

        }

    }

    if(on == -)

        for(int i = ;i < len;i++)

            y[i].r /= len;

}

const int M =;          // a数组所有元素+M，使a[i]>=0

const int MAXN = ;

Complex x1[MAXN];

int a[MAXN/];                //原数组

long long num[MAXN];     //利用FFT得到的数组

long long tt[MAXN];        //统计数组每个元素出现个数

int main()

{

    int n=;                   // n表示除了0之外数组元素个数

    int tot;

    scanf("%d",&tot);

    memset(num,,sizeof(num));

    memset(tt,,sizeof(tt));

    int cnt0=;             //cnt0 统计0的个数

    int aa;

    for(int i = ;i < tot;i++)

        {

            scanf("%d",&aa);

            if(aa==) {cnt0++;continue;}  //先把0全删掉，最后特殊考虑0

            else a[n]=aa;

            num[a[n]+M]++;

            tt[a[n]+M]++;

            n++;

        }

    sort(a,a+n);

    int len1 = a[n-]+M+;

    int len = ;

    while( len < *len1 ) len <<= ;

    for(int i = ;i < len1;i++){

         x1[i] = Complex(num[i],);

    }

    for(int i = len1;i < len;i++){

         x1[i] =Complex(,);

    }

    fft(x1,len,);

    for(int i = ;i < len;i++){

        x1[i] = x1[i]*x1[i];

    }

    fft(x1,len,-);

    for(int i = ;i < len;i++){

         num[i] = (long long)(x1[i].r+0.5);

    }

    len = *(a[n-]+M);

    for(int i = ;i < n;i++) //删掉ai+ai的情况

           num[a[i]+a[i]+*M]--;

/*

   for(int i = 0;i < len;i++){

            if(num[i]) cout<<i-2*M<<' '<<num[i]<<endl;

    }

*/

    long long ret= ;

    int l=a[n-]+M;

    for(int i = ;i <=l; i++)   //ai,aj,ak都不为0的情况

        {

            if(tt[i]) ret+=(long long)(num[i+M]*tt[i]);

        }

    ret+=(long long)(num[*M]*cnt0);        // ai+aj=0的情况

    if(cnt0!=)

        {

            if(cnt0>=) {                   //ai,aj,ak都为0的情况

                long long tmp=;

                tmp*=(long long)(cnt0);

                tmp*=(long long)(cnt0-);

                tmp*=(long long)(cnt0-);

                ret+=tmp;

            }

             for(int i = ;i <=l; i++)

                {

                    if(tt[i]>=){             // x+0=x的情况

                        long long tmp=(long long)cnt0;

                        tmp*=(long long)(tt[i]);

                        tmp*=(long long)(tt[i]-);

                        ret+=tmp*;

                    }

                }

        }

    printf("%lld\n",ret);

    return ;

}