#include <bits/stdc++.h>
using namespace std;
int esq[1000010], dir[1000010], v[1000010];
int v1[1000010];
int tree_max[1000010], tree_min[1000010];
long long ans;
void update_max(int u, int pos) {
  while (u < 1000010) {
    tree_max[u] = max(tree_max[u], pos);
    u += (u & -u);
  }
}
void update_min(int u, int pos) {
  while (u < 1000010) {
    tree_min[u] = min(tree_min[u], pos);
    u += (u & -u);
  }
}
int query_max(int u) {
  int ans = -1;
  while (u) {
    ans = max(ans, tree_max[u]);
    u -= (u & -u);
  }
  return ans;
}
int query_min(int u) {
  int ans = 1e9;
  while (u) {
    ans = min(ans, tree_min[u]);
    u -= (u & -u);
  }
  return ans;
}
set<int> S;
vector<int> S1;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  k--;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    S.insert(v[i]);
  }
  for (auto &u : S) S1.push_back(u);
  for (int i = 0; i < n; i++) {
    v1[i] = (lower_bound(S1.begin(), S1.end(), v[i]) - S1.begin()) + 1;
  }
  for (int i = 1; i < 1000010; i++) tree_min[i] = n, tree_max[i] = -1;
  for (int i = 0; i < n; i++) {
    esq[i] = query_max(1000010 - (v1[i] + 1) - 1);
    update_max(1000010 - v1[i] - 1, i);
  }
  for (int i = n - 1; i >= 0; i--) {
    dir[i] = query_min(1000010 - v1[i] - 1);
    update_min(1000010 - v1[i] - 1, i);
  }
  for (int i = 0; i < n; i++) {
    int a = i - esq[i] - 1;
    int b = dir[i] - i - 1;
    long long x = (long long)v[i];
    if (a > b) swap(a, b);
    for (int i = 0; i <= a; i++) {
      int l1 = (a - i + k - 1) / k;
      if (l1 == 0) l1 = 1;
      int l2 = (a + b - i) / k;
      ans += (x * max(0, l2 - l1 + 1)) % 1000000007LL;
      if (ans >= 1000000007LL) ans -= 1000000007LL;
    }
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
int esq[1000010], dir[1000010], v[1000010];
int v1[1000010];
int tree_max[1000010], tree_min[1000010];
long long ans;
void update_max(int u, int pos) {
  while (u < 1000010) {
    tree_max[u] = max(tree_max[u], pos);
    u += (u & -u);
  }
}
void update_min(int u, int pos) {
  while (u < 1000010) {
    tree_min[u] = min(tree_min[u], pos);
    u += (u & -u);
  }
}
int query_max(int u) {
  int ans = -1;
  while (u) {
    ans = max(ans, tree_max[u]);
    u -= (u & -u);
  }
  return ans;
}
int query_min(int u) {
  int ans = 1e9;
  while (u) {
    ans = min(ans, tree_min[u]);
    u -= (u & -u);
  }
  return ans;
}
set<int> S;
vector<int> S1;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  k--;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    S.insert(v[i]);
  }
  for (auto &u : S) S1.push_back(u);
  for (int i = 0; i < n; i++) {
    v1[i] = (lower_bound(S1.begin(), S1.end(), v[i]) - S1.begin()) + 1;
  }
  for (int i = 1; i < 1000010; i++) tree_min[i] = n, tree_max[i] = -1;
  for (int i = 0; i < n; i++) {
    esq[i] = query_max(1000010 - (v1[i] + 1) - 1);
    update_max(1000010 - v1[i] - 1, i);
  }
  for (int i = n - 1; i >= 0; i--) {
    dir[i] = query_min(1000010 - v1[i] - 1);
    update_min(1000010 - v1[i] - 1, i);
  }
  for (int i = 0; i < n; i++) {
    int a = i - esq[i] - 1;
    int b = dir[i] - i - 1;
    long long x = (long long)v[i];
    if (a > b) swap(a, b);
    for (int i = 0; i <= a; i++) {
      int l1 = (a - i + k - 1) / k;
      if (l1 == 0) l1 = 1;
      int l2 = (a + b - i) / k;
      ans += (x * max(0, l2 - l1 + 1)) % 1000000007LL;
      if (ans >= 1000000007LL) ans -= 1000000007LL;
    }
  }
  cout << ans << "\n";
  return 0;
}

```