Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef vector<vector<long long>> VVL;
typedef vector<string> VS;
typedef pair<int, int> P;
typedef tuple<int,int,int> tpl;

#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REVERSE(c) reverse((c).begin(),(c).end())
#define EXIST(m,v) (m).find((v)) != (m).end()
#define LB(a,x) lower_bound((a).begin(), (a).end(), x) - (a).begin()
#define UB(a,x) upper_bound((a).begin(), (a).end(), x) - (a).begin()

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define RFOR(i,a,b) for(int i=(a)-1;i>=(b);--i)
#define RREP(i,n) RFOR(i,n,0)

#define en "\n"

constexpr double EPS = 1e-9;
constexpr double PI  = 3.1415926535897932;
constexpr int INF = 2147483647;
constexpr long long LINF = 1LL<<60;
constexpr long long MOD = 1000000007; // 998244353;

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

void Main(){
    ll N,V; cin >> N >> V;
    VL a(N), b(N);
    REP(i,N) cin >> a[i];
    REP(i,N) cin >> b[i];

    VL v;
    REP(i,N)REP(j,N){
        v.push_back(a[i]+b[j]);
    }
    SORT(v);

    REP(i,N) cin >> a[i];
    REP(i,N) cin >> b[i];

    ll ans = 0;
    REP(i,N)REP(j,N){
        ll s = a[i]+b[j];
        ll l = LB(v,V-s), r = UB(v,V-s);
        ans += r-l;
    }
    cout << ans << en;
    return;
}

int main(void){
    cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);cout<<fixed<<setprecision(15);
    int t=1; //cin>>t;
    REP(_,t) Main();
    return 0;
}