#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, P = 1e9 + 7;
int n, k, a[N], L[N], R[N];
int get_sum(int a, int b) {
  return (1LL * (a + b) * (b - a + 1) / 2 * (k - 1) + b - a + 1) % P;
}
int find_ans(int l, int p) {
  int en = (l - 1) / (k - 1), ans = en;
  int L = 0, R = en + 1;
  while (L + 1 < R) {
    int mid = L + R >> 1;
    p + mid*(k - 1) <= l ? L = mid : R = mid;
  }
  (ans += (1LL * (en - L) * l + 1LL * L * (p - 1) + get_sum(1, L)) % P) %= P;
  L = 0, R = en + 1;
  while (L + 1 < R) {
    int mid = L + R >> 1;
    p > mid*(k - 1) ? L = mid : R = mid;
  }
  (ans -= (1LL * L * p + get_sum(R, en)) % P) %= P;
  return ans;
}
void readInput() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
}
void prep() {
  for (int i = n - 1; ~i; i--) {
    R[i] = i + 1;
    while (R[i] < n && a[i] > a[R[i]]) R[i] = R[R[i]];
  }
  for (int i = 0; i < n; i++) {
    L[i] = i - 1;
    while (~L[i] && a[i] >= a[L[i]]) L[i] = L[L[i]];
  }
}
void solve() {
  int ans = 0;
  for (int i = 0; i < n; i++)
    (ans += 1LL * a[i] * find_ans(R[i] - L[i] - 1, i - L[i]) % P) %= P;
  cout << (ans + P) % P << endl;
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  readInput(), prep(), solve();
  return 0;
}

### 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 = 1e6 + 5, P = 1e9 + 7;
int n, k, a[N], L[N], R[N];
int get_sum(int a, int b) {
  return (1LL * (a + b) * (b - a + 1) / 2 * (k - 1) + b - a + 1) % P;
}
int find_ans(int l, int p) {
  int en = (l - 1) / (k - 1), ans = en;
  int L = 0, R = en + 1;
  while (L + 1 < R) {
    int mid = L + R >> 1;
    p + mid*(k - 1) <= l ? L = mid : R = mid;
  }
  (ans += (1LL * (en - L) * l + 1LL * L * (p - 1) + get_sum(1, L)) % P) %= P;
  L = 0, R = en + 1;
  while (L + 1 < R) {
    int mid = L + R >> 1;
    p > mid*(k - 1) ? L = mid : R = mid;
  }
  (ans -= (1LL * L * p + get_sum(R, en)) % P) %= P;
  return ans;
}
void readInput() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
}
void prep() {
  for (int i = n - 1; ~i; i--) {
    R[i] = i + 1;
    while (R[i] < n && a[i] > a[R[i]]) R[i] = R[R[i]];
  }
  for (int i = 0; i < n; i++) {
    L[i] = i - 1;
    while (~L[i] && a[i] >= a[L[i]]) L[i] = L[L[i]];
  }
}
void solve() {
  int ans = 0;
  for (int i = 0; i < n; i++)
    (ans += 1LL * a[i] * find_ans(R[i] - L[i] - 1, i - L[i]) % P) %= P;
  cout << (ans + P) % P << endl;
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  readInput(), prep(), solve();
  return 0;
}

```