And here goes another problem on arrays. You are given positive integer len and array a which consists of n integers a1, a2, ..., an. Let's introduce two characteristics for the given array.

  * Let's consider an arbitrary interval of the array with length len, starting in position i. Value <image>, is the modular sum on the chosen interval. In other words, the modular sum is the sum of integers on the chosen interval with length len, taken in its absolute value.
  * Value <image> is the optimal sum of the array. In other words, the optimal sum of an array is the maximum of all modular sums on various intervals of array with length len. 



Your task is to calculate the optimal sum of the given array a. However, before you do the calculations, you are allowed to produce no more than k consecutive operations of the following form with this array: one operation means taking an arbitrary number from array ai and multiply it by -1. In other words, no more than k times you are allowed to take an arbitrary number ai from the array and replace it with  - ai. Each number of the array is allowed to choose an arbitrary number of times.

Your task is to calculate the maximum possible optimal sum of the array after at most k operations described above are completed.

Input

The first line contains two integers n, len (1 ≤ len ≤ n ≤ 105) — the number of elements in the array and the length of the chosen subinterval of the array, correspondingly. 

The second line contains a sequence consisting of n integers a1, a2, ..., an (|ai| ≤ 109) — the original array. 

The third line contains a single integer k (0 ≤ k ≤ n) — the maximum allowed number of operations. 

All numbers in lines are separated by a single space.

Output

In a single line print the maximum possible optimal sum after no more than k acceptable operations are fulfilled. 

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

Examples

Input

5 3
0 -2 3 -5 1
2


Output

10


Input

5 2
1 -3 -10 4 1
3


Output

14


Input

3 3
-2 -5 4
1


Output

11
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int INT_INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const long double PI = acos((long double)-1);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
const int MN = 1e5 + 5;
struct bnode {
  int val, prior, sz;
  long long sum = 0;
  bnode *ch[2] = {0, 0};
  bnode(int val) {
    this->val = val;
    this->prior = rand();
    this->sz = 1;
    this->sum = val;
  }
};
struct bbst {
  bnode *rt;
  bbst(int val) { rt = new bnode(val); }
  inline int sz(bnode *&n) {
    if (n) return n->sz;
    return 0;
  }
  inline int size() {
    if (!rt) return 0;
    return sz(rt);
  }
  inline long long gsum(bnode *&n) {
    if (n) return n->sum;
    return 0;
  }
  inline void upd_sz(bnode *&n) {
    if (n) {
      n->sz = sz(n->ch[0]) + sz(n->ch[1]) + 1;
      n->sum = gsum(n->ch[0]) + gsum(n->ch[1]) + n->val;
    }
  }
  void rot(bnode *&n, int b) {
    bnode *tmp = n->ch[b];
    n->ch[b] = tmp->ch[!b];
    tmp->ch[!b] = n;
    upd_sz(n);
    upd_sz(tmp);
    n = tmp;
  }
  long long getSum(bnode *n, int i, long long sum) {
    if (i == 0) return 0;
    if (sz(n->ch[0]) + 1 == i)
      return sum + gsum(n->ch[0]) + n->val;
    else if (sz(n->ch[0]) + 1 < i)
      return getSum(n->ch[1], i - sz(n->ch[0]) - 1,
                    sum + gsum(n->ch[0]) + n->val);
    else
      return getSum(n->ch[0], i, sum);
  }
  int numLess(bnode *n, long long num) {
    if (!n) return 0;
    if (n->val < num) return sz(n->ch[0]) + 1 + numLess(n->ch[1], num);
    return numLess(n->ch[0], num);
  }
  int kth(bnode *n, int i) {
    if (sz(n->ch[0]) + 1 == i)
      return n->val;
    else if (sz(n->ch[0]) + 1 < i)
      return kth(n->ch[1], i - sz(n->ch[0]) - 1);
    else
      return kth(n->ch[0], i);
  }
  void insert(bnode *&curr, bnode *ins) {
    if (!curr)
      curr = ins;
    else {
      int b = ins->val > curr->val;
      insert(curr->ch[b], ins);
      if (curr->ch[b]->prior > curr->prior) rot(curr, b);
      upd_sz(curr);
    }
  }
  void del(bnode *&n, int i) {
    if (!n) return;
    if (n->val == i) {
      if (!n->ch[0] || !n->ch[1])
        n = n->ch[0] ? n->ch[0] : n->ch[1];
      else {
        bool b = n->ch[1]->prior > n->ch[0]->prior;
        rot(n, b);
        del(n->ch[!b], i);
      }
    } else
      del(n->ch[n->val < i], i);
    upd_sz(n);
  }
  inline void printInOrder() {
    for (int i = 1; i <= sz(rt); i++) cout << kth(rt, i) << " ";
    cout << "\n";
  }
};
int N, K, R, A[MN];
long long res = LLONG_MIN;
bbst *pos, *neg;
inline long long calcPos() {
  long long base = (pos ? pos->getSum(pos->rt, pos->size(), 0) : 0);
  long long sub =
      (neg ? neg->size() <= R ? 0 : neg->getSum(neg->rt, neg->size() - R, 0)
           : 0);
  long long add = (neg ? neg->getSum(neg->rt, neg->size(), 0) - sub : 0);
  return base + add - sub;
}
inline long long calcNeg() {
  long long base = (neg ? neg->getSum(neg->rt, neg->size(), 0) : 0);
  long long sub =
      (pos ? pos->size() <= R ? 0 : pos->getSum(pos->rt, pos->size() - R, 0)
           : 0);
  long long add = (pos ? pos->getSum(pos->rt, pos->size(), 0) - sub : 0);
  return base + add - sub;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> K;
  for (int i = 1; i <= N; i++) cin >> A[i];
  cin >> R;
  for (int i = 1; i <= N; i++) {
    if (A[i] < 0) {
      if (!neg)
        neg = new bbst(-A[i]);
      else
        neg->insert(neg->rt, new bnode(-A[i]));
    } else {
      if (!pos)
        pos = new bbst(A[i]);
      else
        pos->insert(pos->rt, new bnode(A[i]));
    }
    if (i >= K) {
      res = max({res, calcNeg(), calcPos()});
      if (A[i - K + 1] < 0)
        neg->del(neg->rt, -A[i - K + 1]);
      else
        pos->del(pos->rt, A[i - K + 1]);
    }
  }
  cout << res << "\n";
}
