#include <bits/stdc++.h>
using namespace std;
const int md = (int)1e9 + 7;
inline void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += md;
}
inline int mul(int a, int b) { return (int)((long long)a * b % md); }
inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline int inv(int a) {
  a %= md;
  if (a < 0) a += md;
  int b = md, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}
const int INV2 = inv(2);
inline int sum_positive(int start, int finish, int step) {
  if (start <= 0) {
    return 0;
  }
  if (finish < 0) {
    finish += abs(finish) / step * step;
  }
  while (finish < 0) {
    finish += step;
  }
  int cnt = (start - finish) / step + 1;
  int value = mul(mul(start + finish, cnt), INV2);
  return value;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> e(n);
  for (int i = 0; i < n; i++) {
    cin >> e[i].first;
    e[i].second = i;
  }
  sort(e.begin(), e.end());
  vector<int> pr(n), ne(n);
  for (int i = 0; i < n; i++) {
    pr[i] = i - 1;
    ne[i] = i + 1;
  }
  const int step = k - 1;
  int ans = 0;
  for (auto &pp : e) {
    int i = pp.second;
    int L = pr[i] + 1;
    int R = ne[i] - 1;
    int len = (R - L + 1);
    if (len >= k) {
      int pos = i - L;
      int xstart = k;
      int xfinish = k + (len - k) / step * step;
      int xcnt = (xfinish - xstart) / step + 1;
      {
        int value = sum_positive(pos + 1 - xstart, pos + 1 - xfinish, step);
        sub(ans, mul(value, pp.first));
      }
      {
        add(ans, mul(mul(pos, xcnt), pp.first));
        int value = sum_positive(xfinish + pos - len, xstart + pos - len, step);
        sub(ans, mul(value, pp.first));
      }
      add(ans, mul(xcnt, pp.first));
    }
    if (pr[i] != -1) {
      ne[pr[i]] = ne[i];
    }
    if (ne[i] != n) {
      pr[ne[i]] = pr[i];
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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 md = (int)1e9 + 7;
inline void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += md;
}
inline int mul(int a, int b) { return (int)((long long)a * b % md); }
inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline int inv(int a) {
  a %= md;
  if (a < 0) a += md;
  int b = md, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}
const int INV2 = inv(2);
inline int sum_positive(int start, int finish, int step) {
  if (start <= 0) {
    return 0;
  }
  if (finish < 0) {
    finish += abs(finish) / step * step;
  }
  while (finish < 0) {
    finish += step;
  }
  int cnt = (start - finish) / step + 1;
  int value = mul(mul(start + finish, cnt), INV2);
  return value;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> e(n);
  for (int i = 0; i < n; i++) {
    cin >> e[i].first;
    e[i].second = i;
  }
  sort(e.begin(), e.end());
  vector<int> pr(n), ne(n);
  for (int i = 0; i < n; i++) {
    pr[i] = i - 1;
    ne[i] = i + 1;
  }
  const int step = k - 1;
  int ans = 0;
  for (auto &pp : e) {
    int i = pp.second;
    int L = pr[i] + 1;
    int R = ne[i] - 1;
    int len = (R - L + 1);
    if (len >= k) {
      int pos = i - L;
      int xstart = k;
      int xfinish = k + (len - k) / step * step;
      int xcnt = (xfinish - xstart) / step + 1;
      {
        int value = sum_positive(pos + 1 - xstart, pos + 1 - xfinish, step);
        sub(ans, mul(value, pp.first));
      }
      {
        add(ans, mul(mul(pos, xcnt), pp.first));
        int value = sum_positive(xfinish + pos - len, xstart + pos - len, step);
        sub(ans, mul(value, pp.first));
      }
      add(ans, mul(xcnt, pp.first));
    }
    if (pr[i] != -1) {
      ne[pr[i]] = ne[i];
    }
    if (ne[i] != n) {
      pr[ne[i]] = pr[i];
    }
  }
  cout << ans << '\n';
  return 0;
}

```