You have 4 bags A, B, C and D each of which includes N coins (there are totally 4N coins). Values of the coins in each bag are ai, bi, ci and di respectively.

Find the number of combinations that result when you choose one coin from each bag (totally 4 coins) in such a way that the total value of the coins is V. You should distinguish the coins in a bag.

Constraints

* 1 ≤ N ≤ 1000
* 1 ≤ ai, bi, ci, di ≤ 1016
* 1 ≤ V ≤ 1016
* All input values are given in integers

Input

The input is given in the following format.


N V
a1 a2 ... aN
b1 b2 ... bN
c1 c2 ... cN
d1 d2 ... dN


Output

Print the number of combinations in a line.

Examples

Input

3 14
3 1 2
4 8 2
1 2 3
7 3 2


Output

9


Input

5 4
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1


Output

625
#include<iostream>
#include<vector>
#include<algorithm>
#include<cctype>
#include<utility>
#include<string>
#include<cmath>
#include<cstring>
#include<queue>
#include<map>

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define llong long long
#define pb(a) push_back(a)
#define INF 1000000000
using namespace std;
typedef pair<int, int> P;
typedef pair<llong, llong> LP;
typedef pair<int, P> PP;
typedef pair<llong, LP> LPP;
typedef long long int ll;

#define ARRAY_MAX 505

void build(vector<ll>& a,vector<ll>& b,vector<ll>& sum){

    REP(i,a.size()){
        REP(j,b.size()){
            sum.push_back(a[i]+b[j]);
        }
    }
}

int main(){

    ll n,v;
    cin >> n >> v;
    vector<ll> a(n),b(n),c(n),d(n);
    REP(i,n){
        cin >> a[i];
    }    
    REP(i,n){
        cin >> b[i];
    }   
    REP(i,n){
        cin >> c[i];
    }   
    REP(i,n){
        cin >> d[i];
    }

    vector<ll> ab,cd;
    build(a,b,ab);
    build(c,d,cd);

    VSORT(cd);

    ll cnt = 0;
    REP(i,ab.size()){

        cnt += upper_bound(cd.begin(),cd.end(),v-ab[i]) - lower_bound(cd.begin(),cd.end(),v-ab[i]);
    }
    cout << cnt << endl;


    return 0;
}
