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 <cstdio>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef vector<ll>::iterator IT;
const int M = 1024;

ll a[4][M];

int main()
{
	int n; ll w, ans = 0;
	scanf("%d %lld", &n, &w);
	for (int i = 0; i < 4; ++i)
		for (int j = 0; j < n; ++j)
			scanf("%lld", &a[i][j]);
	
	vector<ll> b, c;
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j)
		{
			b.push_back(a[0][i] + a[1][j]);
			c.push_back(a[2][i] + a[3][j]);
		}
	sort(b.begin(), b.end());
	sort(c.begin(), c.end());
	
	
	for (IT it= b.begin(); it != b.end(); ++it)// 加的长度为 大于 - 正好等于的第一个 
		ans += upper_bound(c.begin(), c.end(), w-(*it)) - lower_bound(c.begin(), c.end(), w-(*it));
	printf("%lld\n", ans);
	
	return 0;
}