#include <bits/stdc++.h>
using namespace std;
const int N = 1000050;
const long long Mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
int n, k, arr[N], sleft[N], sright[N], rnd;
vector<int> vec;
stack<pair<int, int> > S;
long long getMod(long long x) { return (x % Mod + Mod) % Mod; }
long long getNumber(long long l, long long r) {
  long long v = k - 1;
  long long a = n / v;
  long long b = l / v + 1ll;
  long long s1 = (v + a * v) * a / 2;
  long long c = v * b - l;
  long long d = a - b + 1ll;
  if (d > 0) {
    long long s2 = (2ll * c + (d - 1ll) * v) * d / 2;
    s1 -= s2;
  }
  s1 += a;
  long long e = r / v + 1ll;
  long long f = e * v - r;
  f = min(f, 1ll + l);
  long long g = (l + 1ll - f) / v + 1ll;
  g = min(g, a - e + 1ll);
  long long s3 = (f + f + (g - 1ll) * v) * g / 2;
  s1 -= s3;
  s1 -= (a - e + 1ll - g) * (l + 1ll);
  return getMod(s1);
}
long long cal(int l, int r) {
  if (l > r) return 0;
  if (r - l + 1 - k < 0) return 0;
  int m = (r - l + 2 - k) / (k - 1);
  long long ret = 1ll * (m + 1) * (r - l + 2 - k) % Mod -
                  1ll * (1 + m) * m / 2 % Mod * (k - 1) % Mod;
  return ret;
}
int main(int argc, const char* argv[]) {
  scanf("%d%d", &n, &k);
  memset(arr, 0, sizeof arr);
  arr[0] = inf;
  arr[n + 1] = inf;
  for (int i = 1; i < ((n) + 1); i++) scanf("%d", arr + i);
  for (int i = 1; i <= n; i++) {
    if (i == 1)
      S.push(make_pair(arr[1], 1)), sleft[1] = 1;
    else {
      sleft[i] = i;
      while (S.size() && S.top().first < arr[i])
        sleft[i] = S.top().second, S.pop();
      S.push(make_pair(arr[i], sleft[i]));
    }
  }
  while (S.size()) S.pop();
  for (int i = n; i >= 1; i--) {
    if (i == n)
      S.push(make_pair(arr[n], n)), sright[n] = n;
    else {
      sright[i] = i;
      while (S.size() && S.top().first <= arr[i])
        sright[i] = S.top().second, S.pop();
      S.push(make_pair(arr[i], sright[i]));
    }
  }
  long long res = 0;
  for (int i = 1; i < ((n) + 1); i++) {
    long long num =
        cal(sleft[i], sright[i]) - cal(sleft[i], i - 1) - cal(i + 1, sright[i]);
    res = getMod(res + getMod(num * arr[i]));
  }
  cout << res << endl;
  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 N = 1000050;
const long long Mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
int n, k, arr[N], sleft[N], sright[N], rnd;
vector<int> vec;
stack<pair<int, int> > S;
long long getMod(long long x) { return (x % Mod + Mod) % Mod; }
long long getNumber(long long l, long long r) {
  long long v = k - 1;
  long long a = n / v;
  long long b = l / v + 1ll;
  long long s1 = (v + a * v) * a / 2;
  long long c = v * b - l;
  long long d = a - b + 1ll;
  if (d > 0) {
    long long s2 = (2ll * c + (d - 1ll) * v) * d / 2;
    s1 -= s2;
  }
  s1 += a;
  long long e = r / v + 1ll;
  long long f = e * v - r;
  f = min(f, 1ll + l);
  long long g = (l + 1ll - f) / v + 1ll;
  g = min(g, a - e + 1ll);
  long long s3 = (f + f + (g - 1ll) * v) * g / 2;
  s1 -= s3;
  s1 -= (a - e + 1ll - g) * (l + 1ll);
  return getMod(s1);
}
long long cal(int l, int r) {
  if (l > r) return 0;
  if (r - l + 1 - k < 0) return 0;
  int m = (r - l + 2 - k) / (k - 1);
  long long ret = 1ll * (m + 1) * (r - l + 2 - k) % Mod -
                  1ll * (1 + m) * m / 2 % Mod * (k - 1) % Mod;
  return ret;
}
int main(int argc, const char* argv[]) {
  scanf("%d%d", &n, &k);
  memset(arr, 0, sizeof arr);
  arr[0] = inf;
  arr[n + 1] = inf;
  for (int i = 1; i < ((n) + 1); i++) scanf("%d", arr + i);
  for (int i = 1; i <= n; i++) {
    if (i == 1)
      S.push(make_pair(arr[1], 1)), sleft[1] = 1;
    else {
      sleft[i] = i;
      while (S.size() && S.top().first < arr[i])
        sleft[i] = S.top().second, S.pop();
      S.push(make_pair(arr[i], sleft[i]));
    }
  }
  while (S.size()) S.pop();
  for (int i = n; i >= 1; i--) {
    if (i == n)
      S.push(make_pair(arr[n], n)), sright[n] = n;
    else {
      sright[i] = i;
      while (S.size() && S.top().first <= arr[i])
        sright[i] = S.top().second, S.pop();
      S.push(make_pair(arr[i], sright[i]));
    }
  }
  long long res = 0;
  for (int i = 1; i < ((n) + 1); i++) {
    long long num =
        cal(sleft[i], sright[i]) - cal(sleft[i], i - 1) - cal(i + 1, sright[i]);
    res = getMod(res + getMod(num * arr[i]));
  }
  cout << res << endl;
  return 0;
}

```