#include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 30;
const int mod = 1e9 + 7;
const int MAXN = 1e6 + 10;
int n, k, ans;
int inv2;
int L[MAXN], R[MAXN], da[MAXN];
int fpow(int a, int b) {
  int ret = 1, tmp = a;
  while (b) {
    if (b & 1) ret = ret * 1ll * tmp % mod;
    tmp = tmp * 1ll * tmp % mod;
    b >>= 1;
  }
  return ret;
}
void get() {
  da[0] = da[n + 1] = inf;
  stack<int> stk;
  stk.push(0);
  for (int i = (1), _END_ = (n); i <= _END_; i++) {
    while (stk.size() && da[i] > da[stk.top()]) stk.pop();
    L[i] = stk.top();
    stk.push(i);
  }
  while (!stk.empty()) stk.pop();
  stk.push(n + 1);
  for (int i = (n), _END_ = (1); i >= _END_; i--) {
    while (stk.size() && da[i] >= da[stk.top()]) stk.pop();
    R[i] = stk.top();
    stk.push(i);
  }
}
int cal(int len) {
  if (k > len) return 0;
  int ret = len - k + 1;
  int num = (len - k) / (k - 1);
  int sum = 0;
  (ret += 1ll * (len + 1) * num % mod) %= mod;
  sum = 1ll * (mod + 1ll * k * (num + 3) % mod - num - 1) % mod * num % mod *
        inv2 % mod;
  return (mod + ret - sum) % mod;
}
int cover(int nl, int nr) {
  int add = cal(nl + nr + 1);
  int mis = (cal(nl) + cal(nr)) % mod;
  return (mod + add - mis) % mod;
}
void init() {
  scanf("%d%d", &n, &k);
  for (int i = (1), _END_ = (n); i <= _END_; i++) scanf("%d", da + i);
  get();
  inv2 = fpow(2, mod - 2);
}
void solve() {
  int num;
  for (int i = (1), _END_ = (n); i <= _END_; i++) {
    num = cover(i - L[i] - 1, R[i] - i - 1);
    (ans += 1ll * da[i] * num % mod) %= mod;
  }
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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 inf = 1 << 30;
const int mod = 1e9 + 7;
const int MAXN = 1e6 + 10;
int n, k, ans;
int inv2;
int L[MAXN], R[MAXN], da[MAXN];
int fpow(int a, int b) {
  int ret = 1, tmp = a;
  while (b) {
    if (b & 1) ret = ret * 1ll * tmp % mod;
    tmp = tmp * 1ll * tmp % mod;
    b >>= 1;
  }
  return ret;
}
void get() {
  da[0] = da[n + 1] = inf;
  stack<int> stk;
  stk.push(0);
  for (int i = (1), _END_ = (n); i <= _END_; i++) {
    while (stk.size() && da[i] > da[stk.top()]) stk.pop();
    L[i] = stk.top();
    stk.push(i);
  }
  while (!stk.empty()) stk.pop();
  stk.push(n + 1);
  for (int i = (n), _END_ = (1); i >= _END_; i--) {
    while (stk.size() && da[i] >= da[stk.top()]) stk.pop();
    R[i] = stk.top();
    stk.push(i);
  }
}
int cal(int len) {
  if (k > len) return 0;
  int ret = len - k + 1;
  int num = (len - k) / (k - 1);
  int sum = 0;
  (ret += 1ll * (len + 1) * num % mod) %= mod;
  sum = 1ll * (mod + 1ll * k * (num + 3) % mod - num - 1) % mod * num % mod *
        inv2 % mod;
  return (mod + ret - sum) % mod;
}
int cover(int nl, int nr) {
  int add = cal(nl + nr + 1);
  int mis = (cal(nl) + cal(nr)) % mod;
  return (mod + add - mis) % mod;
}
void init() {
  scanf("%d%d", &n, &k);
  for (int i = (1), _END_ = (n); i <= _END_; i++) scanf("%d", da + i);
  get();
  inv2 = fpow(2, mod - 2);
}
void solve() {
  int num;
  for (int i = (1), _END_ = (n); i <= _END_; i++) {
    num = cover(i - L[i] - 1, R[i] - i - 1);
    (ans += 1ll * da[i] * num % mod) %= mod;
  }
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}

```