#include <bits/stdc++.h>
using namespace std;
const int MAX = 1001000;
int zn[MAX];
int leftBig[MAX];
int rightBig[MAX];
void solve() {
  long long result = 0, base = 1e9 + 7;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> zn[i];
  }
  vector<pair<int, int> > st;
  st.push_back({int(1e9 + 10000), -1});
  for (int i = 0; i < n; ++i) {
    while (zn[i] > st.back().first) {
      st.pop_back();
    }
    leftBig[i] = st.back().second;
    st.push_back({zn[i], i});
  }
  st.clear();
  st.push_back({int(1e9 + 10000), n});
  for (int i = n - 1; i >= 0; --i) {
    while (zn[i] >= st.back().first) {
      st.pop_back();
    }
    rightBig[i] = st.back().second;
    st.push_back({zn[i], i});
  }
  for (int i = 0; i < n; ++i) {
    int a = i - leftBig[i];
    int b = rightBig[i] - i;
    if (a > b) {
      swap(a, b);
    }
    for (int j = 0; j < a; ++j) {
      int d1 = a - j - 1 - 1;
      int d2 = a - j - 1 + b - 1;
      d1 = d1 / (k - 1);
      if (d1 < 0) d1 = 0;
      d2 = d2 / (k - 1);
      if (d2 < 0) d2 = 0;
      long long t = d2 - d1;
      result = (result + t * zn[i]) % base;
    }
  }
  cout << result << endl;
}
int main() {
  const int inId = 0;
  const int outId = 0;
  if (inId > 0) {
    cerr << "redirect stdin to input " << inId << endl;
    string s = "input";
    s += to_string(inId);
    s += ".txt";
    auto r = freopen(s.c_str(), "r", stdin);
    if (r == nullptr) {
      cerr << "invalid input file " << s;
    }
  }
  if (outId > 0) {
    cerr << "redirect stdout to stdout " << outId << endl;
    string s = "stdout";
    s += to_string(outId);
    s += ".txt";
    auto r = freopen(s.c_str(), "w", stdout);
    if (r == nullptr) {
      cerr << "invalid output file " << s;
    }
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  solve();
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAX = 1001000;
int zn[MAX];
int leftBig[MAX];
int rightBig[MAX];
void solve() {
  long long result = 0, base = 1e9 + 7;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> zn[i];
  }
  vector<pair<int, int> > st;
  st.push_back({int(1e9 + 10000), -1});
  for (int i = 0; i < n; ++i) {
    while (zn[i] > st.back().first) {
      st.pop_back();
    }
    leftBig[i] = st.back().second;
    st.push_back({zn[i], i});
  }
  st.clear();
  st.push_back({int(1e9 + 10000), n});
  for (int i = n - 1; i >= 0; --i) {
    while (zn[i] >= st.back().first) {
      st.pop_back();
    }
    rightBig[i] = st.back().second;
    st.push_back({zn[i], i});
  }
  for (int i = 0; i < n; ++i) {
    int a = i - leftBig[i];
    int b = rightBig[i] - i;
    if (a > b) {
      swap(a, b);
    }
    for (int j = 0; j < a; ++j) {
      int d1 = a - j - 1 - 1;
      int d2 = a - j - 1 + b - 1;
      d1 = d1 / (k - 1);
      if (d1 < 0) d1 = 0;
      d2 = d2 / (k - 1);
      if (d2 < 0) d2 = 0;
      long long t = d2 - d1;
      result = (result + t * zn[i]) % base;
    }
  }
  cout << result << endl;
}
int main() {
  const int inId = 0;
  const int outId = 0;
  if (inId > 0) {
    cerr << "redirect stdin to input " << inId << endl;
    string s = "input";
    s += to_string(inId);
    s += ".txt";
    auto r = freopen(s.c_str(), "r", stdin);
    if (r == nullptr) {
      cerr << "invalid input file " << s;
    }
  }
  if (outId > 0) {
    cerr << "redirect stdout to stdout " << outId << endl;
    string s = "stdout";
    s += to_string(outId);
    s += ".txt";
    auto r = freopen(s.c_str(), "w", stdout);
    if (r == nullptr) {
      cerr << "invalid output file " << s;
    }
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  solve();
}

```