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 <bits/stdc++.h>
                                        #include<iostream>
                                        #include<cstdio>
                                        #include<vector>
                                        #include<queue>
                                        #include<map>
                                        #include<cstring>
                                        #include<string>
                                        #include <math.h>
                                        #include<algorithm>
                                    //    #include <boost/multiprecision/cpp_int.hpp>
                                        #include<functional>
                                #define int long long
                                        #define inf  1000000007
                                        #define pa pair<int,int>
                                        #define ll long long
                                        #define pal pair<double,pa>
                                        #define ppa pair<int,int>
                                        #define ppap pair<pa,pa>
                                        #define ssa pair<string,int>
                                        #define  mp make_pair
                                        #define  pb push_back
                                        #define EPS (1e-10)
                                        #define equals(a,b) (fabs((a)-(b))<EPS)
                                  #define VI vector<int>
                                        using namespace std;
                                        
                                  class pas{
                                    	public:
                                    	int x,y;
                                    	pas(int x=0,int y=0):x(x),y(y) {}
                                    	pas operator + (pas p) {return pas(x+p.x,y+p.y);}
                                    	pas operator - (pas p) {return pas(x-p.x,y-p.y);}
                                    	pas operator * (int a) {return pas(x*a,y*a);}
                                    	pas operator / (int a) {return pas(x/a,y/a);}
                                    //	double absv() {return sqrt(norm());}
                                    	int norm() {return x*x+y*y;}
                                    	bool operator < (const pas &p) const{
                                    		return x != p.x ? x<p.x: y<p.y;
                                    	}
                                    	bool operator == (const pas &p) const{
                                    		return x==p.x && y==p.y;
                                    	}
                                    };
int n,V,W;
vector<int> ve1,ve2;

//int dp[2][10001]={0};
int a[4][1010] ;
           signed main(){
      cin>>n>>V;
           	for(int i=0;i<4;i++)for(int j=0;j<n;j++)cin>>a[i][j];
           	
           	for(int i=0;i<n;i++)for(int j=0;j<n;j++)ve1.pb(a[0][i]+a[1][j]);
           	for(int i=0;i<n;i++)for(int j=0;j<n;j++)ve2.pb(a[2][i]+a[3][j]);
           	sort(ve2.begin(),ve2.end());
           	int ans=0;
           	for(auto v:ve1){
           		ans+=upper_bound(ve2.begin(),ve2.end(),V-v)-lower_bound(ve2.begin(),ve2.end(),V-v);
           	}
           	cout<<ans<<endl;
           }
             