#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';) x = x * 10 + c - '0';
  return x;
}
int a[1000000 + 5], t[1048576 * 2 + 5], z[1000000 + 5], zn, s[1000000 + 5],
    s1[1000000 + 5], s2[1000000 + 5];
int query(int l, int r) {
  int res = 0;
  for (l += 1048576 - 1, r += 1048576 + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) res = max(res, t[l + 1]);
    if (r & 1) res = max(res, t[r - 1]);
  }
  return res;
}
int main() {
  int n, k, i, j, x, y, ans = 0;
  n = read();
  k = read() - 1;
  for (i = 1; i <= n; ++i) t[i + 1048576] = a[i] = read();
  for (i = 1048576; --i;) t[i] = max(t[i << 1], t[i << 1 | 1]);
  for (i = 1; i <= k; ++i) {
    for (zn = 0, j = i; j + k <= n; j += k) {
      x = query(j, j + k - 1);
      y = 1;
      while (zn && x > z[zn]) y += s[zn--];
      z[++zn] = x;
      s[zn] = y;
      s1[zn] = s1[zn - 1] + s[zn];
      s2[zn] = (s2[zn - 1] + 1LL * x * s[zn]) % 1000000007;
      int l = 1, r = zn, mid, res = 0;
      while (l <= r) {
        mid = l + r >> 1;
        if (z[mid] > a[j + k])
          res = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      ans = (ans + s2[res] + 1LL * (s1[zn] - s1[res]) * a[j + k]) % 1000000007;
    }
  }
  printf("%d", ans);
}

### 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;
inline int read() {
  int x;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';) x = x * 10 + c - '0';
  return x;
}
int a[1000000 + 5], t[1048576 * 2 + 5], z[1000000 + 5], zn, s[1000000 + 5],
    s1[1000000 + 5], s2[1000000 + 5];
int query(int l, int r) {
  int res = 0;
  for (l += 1048576 - 1, r += 1048576 + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) res = max(res, t[l + 1]);
    if (r & 1) res = max(res, t[r - 1]);
  }
  return res;
}
int main() {
  int n, k, i, j, x, y, ans = 0;
  n = read();
  k = read() - 1;
  for (i = 1; i <= n; ++i) t[i + 1048576] = a[i] = read();
  for (i = 1048576; --i;) t[i] = max(t[i << 1], t[i << 1 | 1]);
  for (i = 1; i <= k; ++i) {
    for (zn = 0, j = i; j + k <= n; j += k) {
      x = query(j, j + k - 1);
      y = 1;
      while (zn && x > z[zn]) y += s[zn--];
      z[++zn] = x;
      s[zn] = y;
      s1[zn] = s1[zn - 1] + s[zn];
      s2[zn] = (s2[zn - 1] + 1LL * x * s[zn]) % 1000000007;
      int l = 1, r = zn, mid, res = 0;
      while (l <= r) {
        mid = l + r >> 1;
        if (z[mid] > a[j + k])
          res = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      ans = (ans + s2[res] + 1LL * (s1[zn] - s1[res]) * a[j + k]) % 1000000007;
    }
  }
  printf("%d", ans);
}

```