#include <bits/stdc++.h>
using namespace std;
const int p = 1000000007;
int n, k, a[1000005], l[1000005], r[1000005];
stack<int> sta;
int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
long long calc(int x) {
  int L = l[x] + 1, R = r[x] - 1;
  long long ans = 0, y, z, u;
  if (R - x > x - L) {
    y = (int)floor((x - L) * 1.0 / (k - 1));
    z = (int)ceil((R - x) * 1.0 / (k - 1));
    u = (int)floor((R - L) * 1.0 / (k - 1));
    ans += (z - y - 1) * (x - L + 1);
    ans %= p;
  } else {
    y = (int)floor((R - x) * 1.0 / (k - 1));
    z = (int)ceil((x - L) * 1.0 / (k - 1));
    u = (int)floor((R - L) * 1.0 / (k - 1));
    ans += (z - y - 1) * (R - x + 1);
    ans %= p;
  }
  ans += y * (y + 1) / 2 % p * (k - 1) + y;
  ans %= p;
  ans +=
      (R - L + 1) * (u - z + 1) % p - (z + u) * (u - z + 1) / 2 % p * (k - 1);
  ans %= p;
  return ans;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) {
    while (!sta.empty() && a[sta.top()] < a[i]) {
      r[sta.top()] = i;
      sta.pop();
    }
    sta.push(i);
  }
  while (!sta.empty()) {
    r[sta.top()] = n + 1;
    sta.pop();
  }
  for (int i = n; i >= 1; i--) {
    while (!sta.empty() && a[sta.top()] <= a[i]) {
      l[sta.top()] = i;
      sta.pop();
    }
    sta.push(i);
  }
  while (!sta.empty()) {
    l[sta.top()] = 0;
    sta.pop();
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += a[i] * calc(i);
    ans %= p;
  }
  printf("%I64d\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 p = 1000000007;
int n, k, a[1000005], l[1000005], r[1000005];
stack<int> sta;
int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
long long calc(int x) {
  int L = l[x] + 1, R = r[x] - 1;
  long long ans = 0, y, z, u;
  if (R - x > x - L) {
    y = (int)floor((x - L) * 1.0 / (k - 1));
    z = (int)ceil((R - x) * 1.0 / (k - 1));
    u = (int)floor((R - L) * 1.0 / (k - 1));
    ans += (z - y - 1) * (x - L + 1);
    ans %= p;
  } else {
    y = (int)floor((R - x) * 1.0 / (k - 1));
    z = (int)ceil((x - L) * 1.0 / (k - 1));
    u = (int)floor((R - L) * 1.0 / (k - 1));
    ans += (z - y - 1) * (R - x + 1);
    ans %= p;
  }
  ans += y * (y + 1) / 2 % p * (k - 1) + y;
  ans %= p;
  ans +=
      (R - L + 1) * (u - z + 1) % p - (z + u) * (u - z + 1) / 2 % p * (k - 1);
  ans %= p;
  return ans;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) {
    while (!sta.empty() && a[sta.top()] < a[i]) {
      r[sta.top()] = i;
      sta.pop();
    }
    sta.push(i);
  }
  while (!sta.empty()) {
    r[sta.top()] = n + 1;
    sta.pop();
  }
  for (int i = n; i >= 1; i--) {
    while (!sta.empty() && a[sta.top()] <= a[i]) {
      l[sta.top()] = i;
      sta.pop();
    }
    sta.push(i);
  }
  while (!sta.empty()) {
    l[sta.top()] = 0;
    sta.pop();
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += a[i] * calc(i);
    ans %= p;
  }
  printf("%I64d\n", ans);
  return 0;
}

```