#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const long long Mod = 1e9 + 7;
int n, k;
long long ans;
struct node {
  long long v;
  int p;
  node() {}
  node(int a, long long b) : p(a), v(b) {}
} stk[N];
int L[N], R[N], top;
long long times[N], V[N];
long long calc(int t) {
  if (t < k) return t;
  if (times[t]) return t + times[t];
  long long now = t + calc(t - k + 1);
  return times[t] = now;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = k; i <= n; i++) times[i] = calc(i - k + 1) % Mod;
  for (int i = 1; i <= n; i++) scanf("%I64d", &V[i]);
  for (int i = 1; i <= n; i++) {
    while (top > 0 && stk[top].v <= V[i]) --top;
    if (!top) {
      L[i] = 1;
    } else {
      L[i] = stk[top].p + 1;
    }
    stk[++top] = node(i, V[i]);
  }
  top = 0;
  for (int i = n; i >= 1; i--) {
    while (top > 0 && stk[top].v < V[i]) --top;
    if (!top) {
      R[i] = n;
    } else {
      R[i] = stk[top].p - 1;
    }
    stk[++top] = node(i, V[i]);
  }
  for (int i = 1; i <= n; i++) {
    long long now = R[i] - L[i] + 1;
    if (now + 1 >= k) {
      ans = (ans +
             ((times[now] - times[i - L[i]] - times[R[i] - i]) % Mod + Mod) *
                 V[i] % Mod) %
            Mod;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 + 10;
const long long Mod = 1e9 + 7;
int n, k;
long long ans;
struct node {
  long long v;
  int p;
  node() {}
  node(int a, long long b) : p(a), v(b) {}
} stk[N];
int L[N], R[N], top;
long long times[N], V[N];
long long calc(int t) {
  if (t < k) return t;
  if (times[t]) return t + times[t];
  long long now = t + calc(t - k + 1);
  return times[t] = now;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = k; i <= n; i++) times[i] = calc(i - k + 1) % Mod;
  for (int i = 1; i <= n; i++) scanf("%I64d", &V[i]);
  for (int i = 1; i <= n; i++) {
    while (top > 0 && stk[top].v <= V[i]) --top;
    if (!top) {
      L[i] = 1;
    } else {
      L[i] = stk[top].p + 1;
    }
    stk[++top] = node(i, V[i]);
  }
  top = 0;
  for (int i = n; i >= 1; i--) {
    while (top > 0 && stk[top].v < V[i]) --top;
    if (!top) {
      R[i] = n;
    } else {
      R[i] = stk[top].p - 1;
    }
    stk[++top] = node(i, V[i]);
  }
  for (int i = 1; i <= n; i++) {
    long long now = R[i] - L[i] + 1;
    if (now + 1 >= k) {
      ans = (ans +
             ((times[now] - times[i - L[i]] - times[R[i] - i]) % Mod + Mod) *
                 V[i] % Mod) %
            Mod;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```