#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * inf;
const int N = 1000000 + 7;
const int M = 10000 + 7;
const int multipleTest = 0;
const double pi = acos(-1);
template <typename T>
void addmod(T& a, const long long& b, const int MOD = inf) {
  a = (a + b) % inf;
  if (a < 0) a += inf;
}
int n, k;
int a[N];
vector<pair<int, int> > h;
set<int> present;
long long dp[N];
void solve() {
  cin >> n >> k;
  for (int i = (0), _b = (n); i < _b; ++i) {
    scanf("%d", a + i);
    h.push_back({-a[i], i + 1});
  }
  sort(h.begin(), h.end());
  dp[0] = 0;
  for (int i = 1; i <= n; ++i) {
    long long t = i / (k - 1);
    addmod(dp[i], i * t - t * (t + 1) / 2 * (k - 1));
  }
  long long res = 0;
  for (auto& p : h) {
    long long u = p.first;
    int i = p.second;
    int lhs = 1, rhs = n;
    auto it = present.lower_bound(i);
    if (it != present.end()) rhs = (*it) - 1;
    if (it != present.begin()) {
      --it;
      lhs = (*it) + 1;
    }
    long long cur = 0;
    addmod(cur, dp[rhs - lhs + 1] - dp[i - lhs] - dp[rhs - i]);
    addmod(res, cur * -u);
    present.insert(i);
  }
  cout << res << '\n';
}
int main() {
  int Test = 1;
  if (multipleTest) {
    cin >> Test;
  }
  for (int i = 0; i < Test; ++i) {
    solve();
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

Given an array a of n integers and an integer k (2 ≤ k ≤ n), where each element of the array is denoted by a_i (0 ≤ i < n). Perform the operation z given below on a and print the value of z(a,k) modulo 10^{9}+7.
    
    
      
    function z(array a, integer k):  
        if length(a) < k:  
            return 0  
        else:  
            b = empty array  
            ans = 0  
            for i = 0 .. (length(a) - k):  
                temp = a[i]  
                for j = i .. (i + k - 1):  
                    temp = max(temp, a[j])  
                append temp to the end of b  
                ans = ans + temp  
            return ans + z(b, k)  
    

Input

The first line of input contains two integers n and k (2 ≤ k ≤ n ≤ 10^6) — the length of the initial array a and the parameter k.

The second line of input contains n integers a_0, a_1, …, a_{n - 1} (1 ≤ a_{i} ≤ 10^9) — the elements of the array a.

Output

Output the only integer, the value of z(a,k) modulo 10^9+7.

Examples

Input

3 2
9 1 10


Output

29


Input

5 3
5 8 7 1 9


Output

34

Note

In the first example: 

  * for a=(9,1,10), ans=19 and b=(9,10), 
  * for a=(9,10), ans=10 and b=(10), 
  * for a=(10), ans=0. 



So the returned value is 19+10+0=29.

In the second example: 

  * for a=(5,8,7,1,9), ans=25 and b=(8,8,9), 
  * for a=(8,8,9), ans=9 and b=(9), 
  * for a=(9), ans=0. 



So the returned value is 25+9+0=34.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * inf;
const int N = 1000000 + 7;
const int M = 10000 + 7;
const int multipleTest = 0;
const double pi = acos(-1);
template <typename T>
void addmod(T& a, const long long& b, const int MOD = inf) {
  a = (a + b) % inf;
  if (a < 0) a += inf;
}
int n, k;
int a[N];
vector<pair<int, int> > h;
set<int> present;
long long dp[N];
void solve() {
  cin >> n >> k;
  for (int i = (0), _b = (n); i < _b; ++i) {
    scanf("%d", a + i);
    h.push_back({-a[i], i + 1});
  }
  sort(h.begin(), h.end());
  dp[0] = 0;
  for (int i = 1; i <= n; ++i) {
    long long t = i / (k - 1);
    addmod(dp[i], i * t - t * (t + 1) / 2 * (k - 1));
  }
  long long res = 0;
  for (auto& p : h) {
    long long u = p.first;
    int i = p.second;
    int lhs = 1, rhs = n;
    auto it = present.lower_bound(i);
    if (it != present.end()) rhs = (*it) - 1;
    if (it != present.begin()) {
      --it;
      lhs = (*it) + 1;
    }
    long long cur = 0;
    addmod(cur, dp[rhs - lhs + 1] - dp[i - lhs] - dp[rhs - i]);
    addmod(res, cur * -u);
    present.insert(i);
  }
  cout << res << '\n';
}
int main() {
  int Test = 1;
  if (multipleTest) {
    cin >> Test;
  }
  for (int i = 0; i < Test; ++i) {
    solve();
  }
}

```