You are given an array a_1, a_2, ... , a_n and two integers m and k.

You can choose some subarray a_l, a_{l+1}, ..., a_{r-1}, a_r. 

The cost of subarray a_l, a_{l+1}, ..., a_{r-1}, a_r is equal to ∑_{i=l}^{r} a_i - k ⌈ (r - l + 1)/(m) ⌉, where ⌈ x ⌉ is the least integer greater than or equal to x. 

The cost of empty subarray is equal to zero.

For example, if m = 3, k = 10 and a = [2, -4, 15, -3, 4, 8, 3], then the cost of some subarrays are:

  * a_3 ... a_3: 15 - k ⌈ 1/3 ⌉ = 15 - 10 = 5; 
  * a_3 ... a_4: (15 - 3) - k ⌈ 2/3 ⌉ = 12 - 10 = 2; 
  * a_3 ... a_5: (15 - 3 + 4) - k ⌈ 3/3 ⌉ = 16 - 10 = 6; 
  * a_3 ... a_6: (15 - 3 + 4 + 8) - k ⌈ 4/3 ⌉ = 24 - 20 = 4; 
  * a_3 ... a_7: (15 - 3 + 4 + 8 + 3) - k ⌈ 5/3 ⌉ = 27 - 20 = 7. 



Your task is to find the maximum cost of some subarray (possibly empty) of array a.

Input

The first line contains three integers n, m, and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 10, 1 ≤ k ≤ 10^9).

The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9).

Output

Print the maximum cost of some subarray of array a.

Examples

Input


7 3 10
2 -4 15 -3 4 8 3


Output


7


Input


5 2 1000
-13 -4 -9 -20 -11


Output


0
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#pragma GCC target("avx,tune=native")
using namespace std;
const long long inf = 0x3f3f3f3f3f3f;
const double eps = 1e-6;
const long long mod = 1000000007;
inline string getstr(string &s, long long l, long long r) {
  string ret = "";
  for (long long i = l; i <= r; i++) ret.push_back(s[i]);
  return ret;
}
long long modpow(long long x, long long y, long long md = mod) {
  if (y == 0) return 1;
  long long ret = modpow(x, y >> 1, md);
  ret = (long long)ret * ret % md;
  if (y & 1) ret = (long long)ret * x % md;
  return ret;
}
long long n, m, k;
long long a[300005];
long long dp[300005][11];
void upd(long long &x, long long y) { x = max(x, y); }
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i <= 300003; i++)
    for (long long j = 0; j < m; j++) dp[i][j] = (j == 0 ? 0ll : -inf);
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < m; j++) {
      upd(dp[i + 1][(j + 1) % m], dp[i][j] + a[i] - k * (j == 0));
    }
  }
  cout << *max_element(dp[1], dp[n + 2]) << endl;
  return 0;
}
