#include <bits/stdc++.h>
using namespace std;
const int N = 1000010, INF = 1e9 + 7, M = 1e9 + 7;
int n, k, a[N], lpos[N], rpos[N], ans;
int HALF(int x) { return x & 1 ? (x + M) >> 1 : x >> 1; }
int dcsl(int ss, int xs, int gc) {
  if (xs <= 0) return 0;
  int mx = ss + (xs - 1) * gc;
  return HALF((long long)(ss + mx) % M * xs % M);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  vector<int> g;
  a[0] = a[n + 1] = INF;
  g.push_back(0);
  for (int i = 1; i <= n; ++i) {
    int ret = 0;
    for (int l = 0, r = g.size() - 1, mid = (l + r) >> 1; l <= r;
         mid = (l + r) >> 1)
      if (a[g[mid]] > a[i])
        ret = mid, l = mid + 1;
      else
        r = mid - 1;
    lpos[i] = g[ret];
    while (a[i] >= a[g.back()]) g.pop_back();
    g.push_back(i);
  }
  g.clear();
  g.push_back(n + 1);
  for (int i = n; i >= 1; --i) {
    int ret = 0;
    for (int l = 0, r = g.size() - 1, mid = (l + r) >> 1; l <= r;
         mid = (l + r) >> 1)
      if (a[g[mid]] >= a[i])
        ret = mid, l = mid + 1;
      else
        r = mid - 1;
    rpos[i] = g[ret];
    while (a[i] >= a[g.back()]) g.pop_back();
    g.push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    int t1 = (i - lpos[i] - 1) / (k - 1), t2 = (rpos[i] - i - 1) / (k - 1),
        ret = 0;
    if (t1 < t2) {
      (ret += dcsl(k, t1, k - 1)) %= M;
      (ret += (long long)(i - lpos[i]) * (t2 - t1) % M) %= M;
      (ret += dcsl(rpos[i] - (k - 1) * (t2 + 1) - lpos[i] - 1,
                   (rpos[i] - lpos[i] - 1) / (k - 1) - t2, 1 - k)) %= M;
    } else {
      (ret += dcsl(k, t2, k - 1)) %= M;
      (ret += (long long)(rpos[i] - i) * (t1 - t2) % M) %= M;
      (ret += dcsl(rpos[i] - (k - 1) * (t1 + 1) - lpos[i] - 1,
                   (rpos[i] - lpos[i] - 1) / (k - 1) - t1, 1 - k)) %= M;
    }
    (ans += (long long)ret * a[i] % M) %= M;
  }
  cout << ans;
}

### 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 N = 1000010, INF = 1e9 + 7, M = 1e9 + 7;
int n, k, a[N], lpos[N], rpos[N], ans;
int HALF(int x) { return x & 1 ? (x + M) >> 1 : x >> 1; }
int dcsl(int ss, int xs, int gc) {
  if (xs <= 0) return 0;
  int mx = ss + (xs - 1) * gc;
  return HALF((long long)(ss + mx) % M * xs % M);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  vector<int> g;
  a[0] = a[n + 1] = INF;
  g.push_back(0);
  for (int i = 1; i <= n; ++i) {
    int ret = 0;
    for (int l = 0, r = g.size() - 1, mid = (l + r) >> 1; l <= r;
         mid = (l + r) >> 1)
      if (a[g[mid]] > a[i])
        ret = mid, l = mid + 1;
      else
        r = mid - 1;
    lpos[i] = g[ret];
    while (a[i] >= a[g.back()]) g.pop_back();
    g.push_back(i);
  }
  g.clear();
  g.push_back(n + 1);
  for (int i = n; i >= 1; --i) {
    int ret = 0;
    for (int l = 0, r = g.size() - 1, mid = (l + r) >> 1; l <= r;
         mid = (l + r) >> 1)
      if (a[g[mid]] >= a[i])
        ret = mid, l = mid + 1;
      else
        r = mid - 1;
    rpos[i] = g[ret];
    while (a[i] >= a[g.back()]) g.pop_back();
    g.push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    int t1 = (i - lpos[i] - 1) / (k - 1), t2 = (rpos[i] - i - 1) / (k - 1),
        ret = 0;
    if (t1 < t2) {
      (ret += dcsl(k, t1, k - 1)) %= M;
      (ret += (long long)(i - lpos[i]) * (t2 - t1) % M) %= M;
      (ret += dcsl(rpos[i] - (k - 1) * (t2 + 1) - lpos[i] - 1,
                   (rpos[i] - lpos[i] - 1) / (k - 1) - t2, 1 - k)) %= M;
    } else {
      (ret += dcsl(k, t2, k - 1)) %= M;
      (ret += (long long)(rpos[i] - i) * (t1 - t2) % M) %= M;
      (ret += dcsl(rpos[i] - (k - 1) * (t1 + 1) - lpos[i] - 1,
                   (rpos[i] - lpos[i] - 1) / (k - 1) - t1, 1 - k)) %= M;
    }
    (ans += (long long)ret * a[i] % M) %= M;
  }
  cout << ans;
}

```