#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010, p = 1000000007;
void dfs(int);
int f[maxn], ans = 0;
int n, k, a[maxn], root, lc[maxn] = {0}, rc[maxn] = {0}, s[maxn], top = 0,
                         L[maxn], R[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, j = 0; i <= n; i++) {
    while (1 + j * (k - 1) <= i) j++;
    f[i] = (f[i - 1] + j - 1) % p;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  s[++top] = root = 1;
  for (int i = 2; i <= n; i++) {
    s[top + 1] = 0;
    while (top && a[i] > a[s[top]]) top--;
    if (top)
      rc[s[top]] = i;
    else
      root = i;
    lc[i] = s[top + 1];
    s[++top] = i;
  }
  dfs(root);
  if (ans < 0) ans += p;
  printf("%d\n", ans);
  return 0;
}
void dfs(int x) {
  if (lc[x]) {
    dfs(lc[x]);
    L[x] = L[lc[x]];
  } else
    L[x] = x;
  if (rc[x]) {
    dfs(rc[x]);
    R[x] = R[rc[x]];
  } else
    R[x] = x;
  int n = R[x] - L[x] + 1, l = x - L[x], r = R[x] - x;
  ans = (ans + (long long)a[x] * (f[n] - f[l] - f[r])) % p;
}

### Prompt

Create a solution in CPP for 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 maxn = 1000010, p = 1000000007;
void dfs(int);
int f[maxn], ans = 0;
int n, k, a[maxn], root, lc[maxn] = {0}, rc[maxn] = {0}, s[maxn], top = 0,
                         L[maxn], R[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, j = 0; i <= n; i++) {
    while (1 + j * (k - 1) <= i) j++;
    f[i] = (f[i - 1] + j - 1) % p;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  s[++top] = root = 1;
  for (int i = 2; i <= n; i++) {
    s[top + 1] = 0;
    while (top && a[i] > a[s[top]]) top--;
    if (top)
      rc[s[top]] = i;
    else
      root = i;
    lc[i] = s[top + 1];
    s[++top] = i;
  }
  dfs(root);
  if (ans < 0) ans += p;
  printf("%d\n", ans);
  return 0;
}
void dfs(int x) {
  if (lc[x]) {
    dfs(lc[x]);
    L[x] = L[lc[x]];
  } else
    L[x] = x;
  if (rc[x]) {
    dfs(rc[x]);
    R[x] = R[rc[x]];
  } else
    R[x] = x;
  int n = R[x] - L[x] + 1, l = x - L[x], r = R[x] - x;
  ans = (ans + (long long)a[x] * (f[n] - f[l] - f[r])) % p;
}

```