#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, k, ed;
int a[N], l[N], r[N], st[N];
long long calc1(int l, int r) {
  if (l > r) return 0;
  return 1LL * (l + r) * ((r - l) / (k - 1) + 1) / 2;
}
long long calc2(int l, int r, int t) {
  if (l > r) return 0;
  return 1LL * ((r - l) / (k - 1) + 1) * t;
}
long long calc3(int l, int r, int len) {
  if (l > r) return 0;
  return 1LL * (len - l + 1 + len - r + 1) * ((r - l) / (k - 1) + 1) / 2;
}
long long calc(int i, int l, int r) {
  int t1 = min(i - l + 1, r - i + 1), t2 = max(i - l + 1, r - i + 1),
      len = r - l + 1;
  int x = (t1 - 1) / (k - 1) * (k - 1) + 1;
  int y = (t2 - 1) / (k - 1) * (k - 1) + 1;
  int z = (len - 1) / (k - 1) * (k - 1) + 1;
  return calc1(k, x) % mod + calc2(x + k - 1, y, t1) % mod +
         calc3(y + k - 1, z, len) % mod;
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    ed = 0;
    st[0] = n + 1;
    a[n + 1] = inf;
    for (int i = n; i >= 1; i--) {
      while (ed >= 1 && a[st[ed]] <= a[i]) ed--;
      r[i] = st[ed] - 1;
      st[++ed] = i;
    }
    ed = 0;
    st[0] = 0;
    a[0] = inf;
    for (int i = 1; i <= n; i++) {
      while (ed >= 1 && a[st[ed]] < a[i]) ed--;
      l[i] = st[ed] + 1;
      st[++ed] = i;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      ans = ans + calc(i, l[i], r[i]) % mod * a[i] % mod;
      ans %= mod;
    }
    cout << ans << endl;
  }
}

### Prompt

In CPP, your task is to solve 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 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, k, ed;
int a[N], l[N], r[N], st[N];
long long calc1(int l, int r) {
  if (l > r) return 0;
  return 1LL * (l + r) * ((r - l) / (k - 1) + 1) / 2;
}
long long calc2(int l, int r, int t) {
  if (l > r) return 0;
  return 1LL * ((r - l) / (k - 1) + 1) * t;
}
long long calc3(int l, int r, int len) {
  if (l > r) return 0;
  return 1LL * (len - l + 1 + len - r + 1) * ((r - l) / (k - 1) + 1) / 2;
}
long long calc(int i, int l, int r) {
  int t1 = min(i - l + 1, r - i + 1), t2 = max(i - l + 1, r - i + 1),
      len = r - l + 1;
  int x = (t1 - 1) / (k - 1) * (k - 1) + 1;
  int y = (t2 - 1) / (k - 1) * (k - 1) + 1;
  int z = (len - 1) / (k - 1) * (k - 1) + 1;
  return calc1(k, x) % mod + calc2(x + k - 1, y, t1) % mod +
         calc3(y + k - 1, z, len) % mod;
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    ed = 0;
    st[0] = n + 1;
    a[n + 1] = inf;
    for (int i = n; i >= 1; i--) {
      while (ed >= 1 && a[st[ed]] <= a[i]) ed--;
      r[i] = st[ed] - 1;
      st[++ed] = i;
    }
    ed = 0;
    st[0] = 0;
    a[0] = inf;
    for (int i = 1; i <= n; i++) {
      while (ed >= 1 && a[st[ed]] < a[i]) ed--;
      l[i] = st[ed] + 1;
      st[++ed] = i;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      ans = ans + calc(i, l[i], r[i]) % mod * a[i] % mod;
      ans %= mod;
    }
    cout << ans << endl;
  }
}

```