#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
const int mod = (int)1e9 + 7;
int a[N];
int stk[N];
int prv[N], nxt[N];
int add(int a, int b) { return (a + b) % mod; }
int sub(int a, int b) { return (a - b + mod) % mod; }
int mul(int a, int b) { return (long long)a * b % mod; }
int fit(int len, int k) {
  long long times = (long long)len / (k - 1);
  long long ret = times * len - (k - 1) * times * (times + 1) / 2;
  ret %= mod;
  return ret;
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  {
    int ptr = 0;
    for (int i = 0; i < n; i++) {
      while (ptr - 1 >= 0 && a[stk[ptr - 1]] <= a[i]) {
        ptr--;
      }
      if (ptr == 0) {
        prv[i] = 0;
      } else {
        prv[i] = stk[ptr - 1] + 1;
      }
      stk[ptr++] = i;
    }
  }
  {
    int ptr = 0;
    for (int i = n - 1; i >= 0; i--) {
      while (ptr - 1 >= 0 && a[stk[ptr - 1]] < a[i]) {
        ptr--;
      }
      if (ptr == 0) {
        nxt[i] = n - 1;
      } else {
        nxt[i] = stk[ptr - 1] - 1;
      }
      stk[ptr++] = i;
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int sum = sub(fit(nxt[i] - prv[i] + 1, k),
                  add(fit(i - prv[i], k), fit(nxt[i] - i, k)));
    ans = add(ans, mul(a[i], sum));
  }
  printf("%d\n", ans);
  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 = 1000010;
const int mod = (int)1e9 + 7;
int a[N];
int stk[N];
int prv[N], nxt[N];
int add(int a, int b) { return (a + b) % mod; }
int sub(int a, int b) { return (a - b + mod) % mod; }
int mul(int a, int b) { return (long long)a * b % mod; }
int fit(int len, int k) {
  long long times = (long long)len / (k - 1);
  long long ret = times * len - (k - 1) * times * (times + 1) / 2;
  ret %= mod;
  return ret;
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  {
    int ptr = 0;
    for (int i = 0; i < n; i++) {
      while (ptr - 1 >= 0 && a[stk[ptr - 1]] <= a[i]) {
        ptr--;
      }
      if (ptr == 0) {
        prv[i] = 0;
      } else {
        prv[i] = stk[ptr - 1] + 1;
      }
      stk[ptr++] = i;
    }
  }
  {
    int ptr = 0;
    for (int i = n - 1; i >= 0; i--) {
      while (ptr - 1 >= 0 && a[stk[ptr - 1]] < a[i]) {
        ptr--;
      }
      if (ptr == 0) {
        nxt[i] = n - 1;
      } else {
        nxt[i] = stk[ptr - 1] - 1;
      }
      stk[ptr++] = i;
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int sum = sub(fit(nxt[i] - prv[i] + 1, k),
                  add(fit(i - prv[i], k), fit(nxt[i] - i, k)));
    ans = add(ans, mul(a[i], sum));
  }
  printf("%d\n", ans);
  return 0;
}

```