Problem: John Doe has four arrays: a, b, k, and p. Each array consists of n integers. Elements of all arrays are indexed starting from 1. Array p is a permutation of integers 1 to n.

John invented a game for his friends and himself. Initially a player is given array a. The player must consecutively execute exactly u operations on a. You are permitted to execute the following operations:

  * Operation 1: For each <image> change ai into <image>. Expression <image> means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in language C++ and Java it is marked as "^", in Pascal — as "xor". 
  * Operation 2: For each <image> change ai into api + r. When this operation is executed, all changes are made at the same time. 



After all u operations are applied, the number of points the player gets is determined by the formula <image>. 

John wants to find out what maximum number of points a player can win in his game. Help him.

Input

The first line contains space-separated integers n, u and r (1 ≤ n, u ≤ 30, 0 ≤ r ≤ 100) — the number of elements in each array, the number of operations and the number that describes one of the operations. 

Each of the next four lines contains n space-separated integers — arrays a, b, k, p. The first line has array a, the second line has array b, the third line has array k and the fourth one has array p. 

It is guaranteed that elements of arrays a and b are positive and do not exceed 104 (1 ≤ ai, bi ≤ 104), elements of array k do not exceed 104 in the absolute value (|k| ≤ 104) and p is a permutation of numbers from 1 to n.

Output

On a single line print number s — the maximum number of points that a player can win in John's game.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. 

Examples

Input

3 2 1
7 7 7
8 8 8
1 2 3
1 3 2


Output

96


Input

2 1 0
1 1
1 1
1 -1
1 2


Output

0

Note

In the first sample you should first apply the operation of the first type, then the operation of the second type.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long Set(long long N, long long pos) { return N = N | (1 << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1 << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1 << pos)); }
void CI(int &_x) { scanf("%d", &_x); }
void CO(long long &_x) { cout << _x; }
template <typename T>
void getarray(T a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
template <typename T>
void prLLIarray(T a[], long long n) {
  for (long long i = 0; i < n - 1; i++) cout << a[i] << " ";
  cout << a[n - 1] << endl;
}
const double EPS = 1e-9;
const long long INF = 0x7f7f7f7f;
long long dr8[8] = {1, -1, 0, 0, 1, -1, -1, 1};
long long dc8[8] = {0, 0, -1, 1, 1, 1, -1, -1};
long long dr4[4] = {0, 0, 1, -1};
long long dc4[4] = {-1, 1, 0, 0};
long long kn8r[8] = {1, 2, 2, 1, -1, -2, -2, -1};
long long kn8c[8] = {2, 1, -1, -2, -2, -1, 1, 2};
vector<long long> a, b, k, p;
vector<long long> V;
long long rec(int u, int r, int last) {
  if (u == 0) {
    long long sum = 0;
    for (int i = 0; i < k.size(); i++) sum += (V[i] * k[i]);
    return sum;
  }
  long long ans = -1ll << 50;
  if (u % 2 == 0) {
    long long sum = 0;
    for (int i = 0; i < k.size(); i++) sum += (V[i] * k[i]);
    ans = max(ans, sum);
  }
  if (last == 1) {
    for (int i = 0; i < V.size(); i++) V[i] = (V[i] ^ b[i]);
    ans = max(ans, rec(u - 1, r, 0));
    for (int i = 0; i < V.size(); i++) V[i] = (V[i] ^ b[i]);
  }
  vector<long long> temp = V;
  for (int i = 0; i < V.size(); i++) V[i] = (temp[p[i]] + r);
  ans = max(ans, rec(u - 1, r, 1));
  V = temp;
  return ans;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n, u, r;
  cin >> n >> u >> r;
  for (long long i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    a.push_back(temp);
  }
  for (long long i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    b.push_back(temp);
  }
  for (long long i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    k.push_back(temp);
  }
  for (long long i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    temp--;
    p.push_back(temp);
  }
  V = a;
  cout << rec(u, r, 1) << "\n";
}