#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
const int MAXM = 5e5 + 10;
const int INF = 2147483647;
const int MOD = 1e9 + 7;
long long int a[MAXN], ans, lmax[MAXN], rmax[MAXN];
int n, k;
stack<long long int> stk;
void getMax(int n) {
  while (!stk.empty()) {
    stk.pop();
  }
  stk.push(1);
  a[n + 1] = LONG_LONG_MAX;
  for (int i = 2; i <= n + 1; ++i) {
    if (a[i] <= a[stk.top()]) {
      stk.push(i);
      lmax[i] = 0;
    } else {
      while (!stk.empty() && a[i] >= a[stk.top()]) {
        rmax[stk.top()] = i - stk.top() - 1;
        lmax[i] = lmax[stk.top()] + i - stk.top();
        stk.pop();
      }
      stk.push(i);
    }
  }
}
void solve(int now) {
  long long int minn = min(lmax[now], rmax[now]),
                maxn = max(lmax[now], rmax[now]);
  long long int nowk = k, nxtk = 1, d = k - 1, sum = 0, num;
  if (minn >= k) {
    nxtk = ((minn + 1 - k) / d) * d + k;
    num = (nxtk - nowk) / d + 1;
    if (nxtk >= nowk) {
      sum += ((nowk + nxtk) * num) / 2;
    }
  }
  if (maxn >= k) {
    nowk = nxtk + d;
    nxtk = ((maxn + 1 - k) / d) * d + k;
    num = (nxtk - nowk) / d + 1;
    if (nxtk >= nowk) {
      sum += (minn + 1) * num;
      sum %= MOD;
    }
  }
  long long int p = minn + maxn + 1;
  if (p >= k) {
    nowk = nxtk + d;
    nxtk = ((p - k) / d) * d + k;
    if (nxtk >= nowk) {
      nowk = p - nowk + 1;
      nxtk = p - nxtk + 1;
      swap(nowk, nxtk);
      num = (nxtk - nowk) / d + 1;
      sum += ((nowk + nxtk) * num) / 2;
    }
  }
  sum %= MOD;
  ans += sum * a[now];
  ans %= MOD;
}
void ceshi() {
  for (int i = 1; i <= n; ++i) {
    i == n ? printf("%I64d\n", lmax[i]) : printf("%I64d ", lmax[i]);
  }
  for (int i = 1; i <= n; ++i) {
    i == n ? printf("%I64d\n", rmax[i]) : printf("%I64d ", rmax[i]);
  }
  printf("\n");
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
  }
  getMax(n);
  for (int i = 1; i <= n; ++i) {
    solve(i);
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 MAXN = 1e6 + 10;
const int MAXM = 5e5 + 10;
const int INF = 2147483647;
const int MOD = 1e9 + 7;
long long int a[MAXN], ans, lmax[MAXN], rmax[MAXN];
int n, k;
stack<long long int> stk;
void getMax(int n) {
  while (!stk.empty()) {
    stk.pop();
  }
  stk.push(1);
  a[n + 1] = LONG_LONG_MAX;
  for (int i = 2; i <= n + 1; ++i) {
    if (a[i] <= a[stk.top()]) {
      stk.push(i);
      lmax[i] = 0;
    } else {
      while (!stk.empty() && a[i] >= a[stk.top()]) {
        rmax[stk.top()] = i - stk.top() - 1;
        lmax[i] = lmax[stk.top()] + i - stk.top();
        stk.pop();
      }
      stk.push(i);
    }
  }
}
void solve(int now) {
  long long int minn = min(lmax[now], rmax[now]),
                maxn = max(lmax[now], rmax[now]);
  long long int nowk = k, nxtk = 1, d = k - 1, sum = 0, num;
  if (minn >= k) {
    nxtk = ((minn + 1 - k) / d) * d + k;
    num = (nxtk - nowk) / d + 1;
    if (nxtk >= nowk) {
      sum += ((nowk + nxtk) * num) / 2;
    }
  }
  if (maxn >= k) {
    nowk = nxtk + d;
    nxtk = ((maxn + 1 - k) / d) * d + k;
    num = (nxtk - nowk) / d + 1;
    if (nxtk >= nowk) {
      sum += (minn + 1) * num;
      sum %= MOD;
    }
  }
  long long int p = minn + maxn + 1;
  if (p >= k) {
    nowk = nxtk + d;
    nxtk = ((p - k) / d) * d + k;
    if (nxtk >= nowk) {
      nowk = p - nowk + 1;
      nxtk = p - nxtk + 1;
      swap(nowk, nxtk);
      num = (nxtk - nowk) / d + 1;
      sum += ((nowk + nxtk) * num) / 2;
    }
  }
  sum %= MOD;
  ans += sum * a[now];
  ans %= MOD;
}
void ceshi() {
  for (int i = 1; i <= n; ++i) {
    i == n ? printf("%I64d\n", lmax[i]) : printf("%I64d ", lmax[i]);
  }
  for (int i = 1; i <= n; ++i) {
    i == n ? printf("%I64d\n", rmax[i]) : printf("%I64d ", rmax[i]);
  }
  printf("\n");
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
  }
  getMax(n);
  for (int i = 1; i <= n; ++i) {
    solve(i);
  }
  printf("%I64d\n", ans);
  return 0;
}

```