#include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
bool Finish_read;
template <class T>
inline void read(T &x) {
  Finish_read = 0;
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    if (ch == EOF) return;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
  Finish_read = 1;
}
template <class T>
inline void print(T x) {
  if (x / 10 != 0) print(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void writeln(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
  putchar('\n');
}
template <class T>
inline void write(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
}
const int maxn = 1000005, mod = 1e9 + 7;
int n, mxl[maxn], mxr[maxn], k;
namespace {
inline int Add(const int &x, const int &y) {
  int res = x + y;
  return res >= mod ? res - mod : res;
}
inline int Sub(const int &x, const int &y) {
  int res = x - y;
  return res < 0 ? res + mod : res;
}
inline int Mul(const int &x, const int &y) { return 1ll * x * y % mod; }
inline int Half(const int &x) { return x & 1 ? x + mod >> 1 : x >> 1; }
}  // namespace
struct xint {
  int val, id;
  inline bool operator<(const xint &rhs) const { return val > rhs.val; }
} v[maxn];
set<int> x;
int ans;
inline void input() {
  read(n), read(k);
  for (int i = 1; i <= n; ++i) read(v[i].val), v[i].id = i;
  sort(v + 1, v + n + 1);
}
int main() {
  input();
  x.insert(0), x.insert(n + 1);
  for (int i = 1; i <= n; ++i) {
    int l = *(--x.lower_bound(v[i].id)), r = *(x.upper_bound(v[i].id)),
        L = max(l + 1, v[i].id - k + 1), R = min(v[i].id, r - k);
    x.insert(v[i].id);
    if (L <= R) {
      int p = r - k - l, q = r - k - v[i].id, r = v[i].id - k - l;
      int tmp = p <= 0
                    ? 0
                    : 1ll *
                          (((long long)p + p - 1ll * (p / (k - 1)) * (k - 1)) *
                           ((p / (k - 1)) + 1) / 2) %
                          mod,
          tmq = q <= 0
                    ? 0
                    : 1ll *
                          (((long long)q + q - 1ll * (q / (k - 1)) * (k - 1)) *
                           ((q / (k - 1)) + 1) / 2) %
                          mod,
          tmr = r <= 0
                    ? 0
                    : 1ll *
                          (((long long)r + r - 1ll * (r / (k - 1)) * (k - 1)) *
                           ((r / (k - 1)) + 1) / 2) %
                          mod;
      ans = Add(ans, Mul(Sub(tmp, Add(tmq, tmr)), v[i].val));
    }
  }
  printf("%d\n", ans);
}

### 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>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
bool Finish_read;
template <class T>
inline void read(T &x) {
  Finish_read = 0;
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    if (ch == EOF) return;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
  Finish_read = 1;
}
template <class T>
inline void print(T x) {
  if (x / 10 != 0) print(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void writeln(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
  putchar('\n');
}
template <class T>
inline void write(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
}
const int maxn = 1000005, mod = 1e9 + 7;
int n, mxl[maxn], mxr[maxn], k;
namespace {
inline int Add(const int &x, const int &y) {
  int res = x + y;
  return res >= mod ? res - mod : res;
}
inline int Sub(const int &x, const int &y) {
  int res = x - y;
  return res < 0 ? res + mod : res;
}
inline int Mul(const int &x, const int &y) { return 1ll * x * y % mod; }
inline int Half(const int &x) { return x & 1 ? x + mod >> 1 : x >> 1; }
}  // namespace
struct xint {
  int val, id;
  inline bool operator<(const xint &rhs) const { return val > rhs.val; }
} v[maxn];
set<int> x;
int ans;
inline void input() {
  read(n), read(k);
  for (int i = 1; i <= n; ++i) read(v[i].val), v[i].id = i;
  sort(v + 1, v + n + 1);
}
int main() {
  input();
  x.insert(0), x.insert(n + 1);
  for (int i = 1; i <= n; ++i) {
    int l = *(--x.lower_bound(v[i].id)), r = *(x.upper_bound(v[i].id)),
        L = max(l + 1, v[i].id - k + 1), R = min(v[i].id, r - k);
    x.insert(v[i].id);
    if (L <= R) {
      int p = r - k - l, q = r - k - v[i].id, r = v[i].id - k - l;
      int tmp = p <= 0
                    ? 0
                    : 1ll *
                          (((long long)p + p - 1ll * (p / (k - 1)) * (k - 1)) *
                           ((p / (k - 1)) + 1) / 2) %
                          mod,
          tmq = q <= 0
                    ? 0
                    : 1ll *
                          (((long long)q + q - 1ll * (q / (k - 1)) * (k - 1)) *
                           ((q / (k - 1)) + 1) / 2) %
                          mod,
          tmr = r <= 0
                    ? 0
                    : 1ll *
                          (((long long)r + r - 1ll * (r / (k - 1)) * (k - 1)) *
                           ((r / (k - 1)) + 1) / 2) %
                          mod;
      ans = Add(ans, Mul(Sub(tmp, Add(tmq, tmr)), v[i].val));
    }
  }
  printf("%d\n", ans);
}

```