#include <iostream>
#include <algorithm>

#define x first
#define y second

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int N = 300010;
PII a[N];
int cnt[3][N];
int n;

int main()
{
    int la, lb, lc, d;
    cin >> la >> lb >> lc >> d;
    for (int i = 1; i <= la; i++)
    {
        int x;
        scanf("%d", &x);
        a[++n] = {x, 0};
    }

    for (int i = 1; i <= lb; i++)
    {
        int x;
        scanf("%d", &x);
        a[++n] = {x, 1};
    }

    for (int i = 1; i <= lc; i++)
    {
        int x;
        scanf("%d", &x);
        a[++n] = {x, 2};
    }

    sort(a + 1, a + n + 1);

    for (int i = 0; i < 3; i++)
        for (int j = 1; j <= n; j++)
            cnt[i][j] = cnt[i][j - 1] + (a[j].y == i);

    int last = -1;
    LL res = 0;
    for (int i = 1, j = 1; i <= n; i++)
    {
        while (j <= n && a[j].x - a[i].x <= d) j++;  //a[j].x - a[i].x > d;

        if (j > last) 
        {
            LL ta = cnt[0][j - 1] - cnt[0][i - 1]; //i ~ j - 1中属于数组A的数有几个
            LL tb = cnt[1][j - 1] - cnt[1][i - 1];
            LL tc = cnt[2][j - 1] - cnt[2][i - 1];
            res += ta * tb * tc;
        }

        if (i <= last && last < j)
        {
            LL ta = cnt[0][last - 1] - cnt[0][i - 1]; 
            LL tb = cnt[1][last - 1] - cnt[1][i - 1];
            LL tc = cnt[2][last - 1] - cnt[2][i - 1];
            res -= ta * tb * tc;
        }

        last = j;

    }

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

    return 0;
}