#include <bits/stdc++.h>
using namespace std;
const int MX = 1000005, mod = 1e9 + 7;
int n, k, q[MX], dp[MX];
long long a[MX], izq[MX], der[MX];
set<int> st;
void obtIzqDer() {
  iota(q, q + n, 0);
  sort(q, q + n, [&](int i, int j) {
    if (a[i] == a[j]) return i < j;
    return a[i] > a[j];
  });
  int i = 0, j = 0;
  st.insert(-1);
  while (i < n) {
    while (j < n && a[q[i]] < a[q[j]]) {
      st.insert(q[j]);
      j++;
    }
    auto it = st.upper_bound(q[i]);
    it--;
    izq[q[i]] = *it + 1;
    i++;
  }
  st.clear();
  st.insert(n);
  i = 0, j = 0;
  while (i < n) {
    while (j < n && a[q[i]] <= a[q[j]]) {
      st.insert(q[j]);
      j++;
    }
    auto it = st.upper_bound(q[i]);
    der[q[i]] = *it - 1;
    i++;
  }
}
long long comb(int a, int b, int c) {
  return (dp[c - a + 1] - dp[b - a] - dp[c - b]) % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = k; i <= n; i++) dp[i] = (dp[i - k + 1] + i - k + 1) % mod;
  obtIzqDer();
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res += a[i] * comb(izq[i], i, der[i]) % mod;
    res %= mod;
  }
  cout << ((((res) % (mod)) + (mod)) % (mod)) << '\n';
  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 MX = 1000005, mod = 1e9 + 7;
int n, k, q[MX], dp[MX];
long long a[MX], izq[MX], der[MX];
set<int> st;
void obtIzqDer() {
  iota(q, q + n, 0);
  sort(q, q + n, [&](int i, int j) {
    if (a[i] == a[j]) return i < j;
    return a[i] > a[j];
  });
  int i = 0, j = 0;
  st.insert(-1);
  while (i < n) {
    while (j < n && a[q[i]] < a[q[j]]) {
      st.insert(q[j]);
      j++;
    }
    auto it = st.upper_bound(q[i]);
    it--;
    izq[q[i]] = *it + 1;
    i++;
  }
  st.clear();
  st.insert(n);
  i = 0, j = 0;
  while (i < n) {
    while (j < n && a[q[i]] <= a[q[j]]) {
      st.insert(q[j]);
      j++;
    }
    auto it = st.upper_bound(q[i]);
    der[q[i]] = *it - 1;
    i++;
  }
}
long long comb(int a, int b, int c) {
  return (dp[c - a + 1] - dp[b - a] - dp[c - b]) % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = k; i <= n; i++) dp[i] = (dp[i - k + 1] + i - k + 1) % mod;
  obtIzqDer();
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res += a[i] * comb(izq[i], i, der[i]) % mod;
    res %= mod;
  }
  cout << ((((res) % (mod)) + (mod)) % (mod)) << '\n';
  return 0;
}

```