Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection. 

Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 ≤ i ≤ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 ≤ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.

Input

The first line of the input contains two space-separated integers n and k (1 ≤ n ≤ 2·k ≤ 100 000), denoting the number of cowbells and the number of boxes, respectively.

The next line contains n space-separated integers s1, s2, ..., sn (1 ≤ s1 ≤ s2 ≤ ... ≤ sn ≤ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.

Output

Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.

Examples

Input

2 1
2 5


Output

7


Input

4 3
2 3 5 9


Output

9


Input

3 2
3 5 7


Output

8

Note

In the first sample, Kevin must pack his two cowbells into the same box. 

In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.

In the third sample, the optimal solution is {3, 5} and {7}.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll md = 1000000007;
ll exp(ll a, ll b) {
  ll r = 1ll;
  while (b > 0) {
    if (b & 1) {
      r = r * (a % md);
      r = (r + md) % md;
    }
    b >>= 1;
    a = (a % md) * (a % md);
    a = (a + md) % md;
  }
  return (r + md) % md;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll Min(ll a, ll b) {
  if (a < b) return a;
  return b;
}
ll Max(ll a, ll b) {
  if (a > b) return a;
  return b;
}
long long n, k;
bool check(long long x, vector<ll>& v) {
  long long used = 0;
  if (x < v.back()) return false;
  long long i = 0, j = n - 1;
  while (i <= j) {
    if (i != j && v[j] + v[i] <= x) ++i;
    --j;
    ++used;
  }
  return used <= k;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  vector<ll> v(n);
  for (ll i = 0; i <= n - 1; ++i) cin >> v[i];
  long long low = 0, high = 1e12, mid, ans = high;
  while (low <= high) {
    mid = (low + high) / 2;
    if (check(mid, v)) {
      ans = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  cout << ans;
  cout << '\n';
  return 0;
}
