#include <bits/stdc++.h>
using namespace std;
double PI = acos(-1);
int const N = 1e6 + 10, M = 2e7 + 1, oo = 1e9;
const int mod = oo + 7;
const long long OO = 3e18;
int n, k, aa[N], lg[N], dp[N][21];
int mxx(int l, int r) {
  int og = lg[r - l + 1];
  return max(dp[l][og], dp[l + r - l + 1 - (1 << og)][og]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  lg[1] = 0;
  for (int i = 2; i < N; i++) lg[i] = lg[i / 2] + 1;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> aa[i], dp[i][0] = aa[i];
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 0; i + (1 << (j - 1)) < n; i++)
      dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
  long long an = 0;
  for (int i = 0; i < k - 1; i++) {
    vector<int> v, lf, rg;
    for (int j = i; j + k - 1 < n; j += k - 1) {
      int mx = mxx(j, j + k - 1);
      v.push_back(mx);
    }
    lf.resize(v.size());
    rg.resize(v.size());
    stack<pair<int, int> > st;
    for (int i = 0; i < (int)v.size(); i++) {
      while (!st.empty() && st.top().first <= v[i]) st.pop();
      if (st.empty())
        lf[i] = -1;
      else
        lf[i] = st.top().second;
      st.push({v[i], i});
    }
    while (!st.empty()) st.pop();
    for (int i = (int)v.size() - 1; i > -1; i--) {
      while (!st.empty() && st.top().first < v[i]) st.pop();
      if (st.empty())
        rg[i] = v.size();
      else
        rg[i] = st.top().second;
      st.push({v[i], i});
    }
    for (int i = 0; i < (int)v.size(); i++) {
      int tol = i - lf[i], tor = rg[i] - i - 1;
      an += (tol * 1LL * tor % mod) * v[i] % mod;
      an %= mod;
      an += tol * 1LL * v[i] % mod;
      an %= mod;
    }
  }
  cout << an;
  return 0;
}

### Prompt

Your task is to create 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;
double PI = acos(-1);
int const N = 1e6 + 10, M = 2e7 + 1, oo = 1e9;
const int mod = oo + 7;
const long long OO = 3e18;
int n, k, aa[N], lg[N], dp[N][21];
int mxx(int l, int r) {
  int og = lg[r - l + 1];
  return max(dp[l][og], dp[l + r - l + 1 - (1 << og)][og]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  lg[1] = 0;
  for (int i = 2; i < N; i++) lg[i] = lg[i / 2] + 1;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> aa[i], dp[i][0] = aa[i];
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 0; i + (1 << (j - 1)) < n; i++)
      dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
  long long an = 0;
  for (int i = 0; i < k - 1; i++) {
    vector<int> v, lf, rg;
    for (int j = i; j + k - 1 < n; j += k - 1) {
      int mx = mxx(j, j + k - 1);
      v.push_back(mx);
    }
    lf.resize(v.size());
    rg.resize(v.size());
    stack<pair<int, int> > st;
    for (int i = 0; i < (int)v.size(); i++) {
      while (!st.empty() && st.top().first <= v[i]) st.pop();
      if (st.empty())
        lf[i] = -1;
      else
        lf[i] = st.top().second;
      st.push({v[i], i});
    }
    while (!st.empty()) st.pop();
    for (int i = (int)v.size() - 1; i > -1; i--) {
      while (!st.empty() && st.top().first < v[i]) st.pop();
      if (st.empty())
        rg[i] = v.size();
      else
        rg[i] = st.top().second;
      st.push({v[i], i});
    }
    for (int i = 0; i < (int)v.size(); i++) {
      int tol = i - lf[i], tor = rg[i] - i - 1;
      an += (tol * 1LL * tor % mod) * v[i] % mod;
      an %= mod;
      an += tol * 1LL * v[i] % mod;
      an %= mod;
    }
  }
  cout << an;
  return 0;
}

```