#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 11, mod = 1e9 + 7;
int n, ans, k;
int L[N], R[N], st[N], a[N];
long long f[N];
inline int calc(int x, int l, int r) {
  return (f[x - l] - (x == r ? 0 : f[x - r - 1])) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (register int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (register int i = 1; i <= n; ++i) {
    while (st[0] && a[st[st[0]]] <= a[i]) R[st[st[0]--]] = i - 1;
    st[++st[0]] = i;
  }
  for (register int i = 1; i <= n; ++i) f[i] = f[i - 1] + i / k;
  for (register int i = 1; i <= st[0]; ++i) R[st[i]] = n;
  st[0] = 0;
  for (register int i = n; i >= 1; --i) {
    while (st[0] && a[st[st[0]]] < a[i]) L[st[st[0]--]] = i + 1;
    st[++st[0]] = i;
  }
  for (register int i = 1; i <= st[0]; ++i) L[st[i]] = 1;
  for (register int i = 1; i <= n; ++i) {
    ans = (ans + 1ll * (i - L[i]) / k * a[i]) % mod;
    ans = (ans + 1ll * (calc(R[i], L[i], i) - calc(i, L[i], i)) * a[i]) % mod;
  }
  ans = (ans + mod) % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
const int N = 1e6 + 11, mod = 1e9 + 7;
int n, ans, k;
int L[N], R[N], st[N], a[N];
long long f[N];
inline int calc(int x, int l, int r) {
  return (f[x - l] - (x == r ? 0 : f[x - r - 1])) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (register int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (register int i = 1; i <= n; ++i) {
    while (st[0] && a[st[st[0]]] <= a[i]) R[st[st[0]--]] = i - 1;
    st[++st[0]] = i;
  }
  for (register int i = 1; i <= n; ++i) f[i] = f[i - 1] + i / k;
  for (register int i = 1; i <= st[0]; ++i) R[st[i]] = n;
  st[0] = 0;
  for (register int i = n; i >= 1; --i) {
    while (st[0] && a[st[st[0]]] < a[i]) L[st[st[0]--]] = i + 1;
    st[++st[0]] = i;
  }
  for (register int i = 1; i <= st[0]; ++i) L[st[i]] = 1;
  for (register int i = 1; i <= n; ++i) {
    ans = (ans + 1ll * (i - L[i]) / k * a[i]) % mod;
    ans = (ans + 1ll * (calc(R[i], L[i], i) - calc(i, L[i], i)) * a[i]) % mod;
  }
  ans = (ans + mod) % mod;
  printf("%d\n", ans);
  return 0;
}

```