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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>

#define REP(i,n) for(int i=0;i<n;++i)
#define REPR(i,n) for(int i=n;i>=0;--i)
#define REPI(itr,v) for(auto itr=v.begin();itr!=v.end();++itr)
#define REPIR(itr,v) for(auto itr=v.rbegin();itr!=v.rend();++itr)
#define FOR(i,a,b) for(int i=a;i<b;++i)
#define SORT(v,n) sort(v, v+n)
#define SORTV(v) sort(v.begin(), v.end())
#define ALL(v) v.begin(),v.end()
#define llong long long
#define INF 999999999
#define SUR 1000000007
#define pb push_back
#define pf push_front
#define MP make_pair

int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};

using namespace std;

typedef pair<int,int> pii;

int main(){

  llong n, v;
  scanf("%lld %lld", &n, &v);
  vector<llong> a(n), b(n), c(n), d(n);
  REP(i,n){
    scanf("%lld", &a[i]);
  }
  REP(i,n){
    scanf("%lld", &b[i]);
  }
  REP(i,n){
    scanf("%lld", &c[i]);
  }
  REP(i,n){
    scanf("%lld", &d[i]);
  }

  map<llong, int> mp;;
  REP(i,n){
    REP(j,n){
      mp[a[i] + b[j]]++;
    }
  }

  llong ans = 0;
  REP(i,n){
    REP(j,n){
      llong tmp = v - c[i] - d[j];
      if(mp[tmp] > 0){
        ans += mp[tmp];
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}


