#include <bits/stdc++.h>
using namespace std;
const int MAXn = 1e6 + 10, MAXlg = 20 + 4, MOD = 1e9 + 7;
int dp1[MAXn][MAXlg], dp2[MAXn][MAXlg], n, k, ar[MAXn], ar2[MAXn], ar3[MAXn];
long long ans;
void make_dp1() {
  for (int i = 1; i < n; i++) dp1[i][0] = ar[i - 1];
  for (int i = 1; i < MAXlg; i++)
    for (int j = 0; j < n; j++)
      if (j - (1 << i) >= 0)
        dp1[j][i] = max(dp1[j][i - 1], dp1[j - (1 << (i - 1))][i - 1]);
}
void make_dp2() {
  for (int i = 0; i < n - 1; i++) dp2[i][0] = ar[i + 1];
  for (int i = 1; i < MAXlg; i++)
    for (int j = 0; j < n; j++)
      if (j + (1 << i) < n)
        dp2[j][i] = max(dp2[j][i - 1], dp2[j + (1 << (i - 1))][i - 1]);
}
int query1(int a, int num) {
  int ansq = 0;
  for (int i = MAXlg - 1; i >= 0; i--) {
    if (a - (1 << i) >= 0 && dp1[a][i] <= num) {
      ansq += (1 << i);
      a -= (1 << i);
    }
  }
  return ansq;
}
int query2(int a, int num) {
  int ansq = 0;
  for (int i = MAXlg - 1; i >= 0; i--) {
    if (a + (1 << i) < n && dp2[a][i] < num) {
      ansq += (1 << i);
      a += (1 << i);
    }
  }
  return ansq;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> ar[i];
  make_dp1();
  make_dp2();
  if (k == 2) {
    for (int i = 2; i < MAXn; i++)
      ar3[i] = ((long long)ar3[i - 1] + i - 1) % MOD;
  } else {
    ar3[k] = 1;
    ar2[k] = 1;
    for (int i = k + 1; i < MAXn; i++) {
      ar2[i] = ar2[i - 1];
      if (i % (k - 1) == 1) ar2[i]++;
      ar2[i] %= MOD;
      ar3[i] = ((long long)ar3[i - 1] + ar2[i]) % MOD;
      ar3[i] %= MOD;
    }
  }
  for (int i = 0; i < n; i++) {
    long long l = query1(i, ar[i]), r = query2(i, ar[i]),
              hp = ((long long)ar3[l + r + 1] - ar3[l] - ar3[r] + MOD) % MOD,
              hp2 = ar[i], ans2 = (hp * hp2) % MOD;
    ans = ((long long)ans + ans2) % MOD;
  }
  cout << ans;
}

### Prompt

Please create a solution in cpp 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;
const int MAXn = 1e6 + 10, MAXlg = 20 + 4, MOD = 1e9 + 7;
int dp1[MAXn][MAXlg], dp2[MAXn][MAXlg], n, k, ar[MAXn], ar2[MAXn], ar3[MAXn];
long long ans;
void make_dp1() {
  for (int i = 1; i < n; i++) dp1[i][0] = ar[i - 1];
  for (int i = 1; i < MAXlg; i++)
    for (int j = 0; j < n; j++)
      if (j - (1 << i) >= 0)
        dp1[j][i] = max(dp1[j][i - 1], dp1[j - (1 << (i - 1))][i - 1]);
}
void make_dp2() {
  for (int i = 0; i < n - 1; i++) dp2[i][0] = ar[i + 1];
  for (int i = 1; i < MAXlg; i++)
    for (int j = 0; j < n; j++)
      if (j + (1 << i) < n)
        dp2[j][i] = max(dp2[j][i - 1], dp2[j + (1 << (i - 1))][i - 1]);
}
int query1(int a, int num) {
  int ansq = 0;
  for (int i = MAXlg - 1; i >= 0; i--) {
    if (a - (1 << i) >= 0 && dp1[a][i] <= num) {
      ansq += (1 << i);
      a -= (1 << i);
    }
  }
  return ansq;
}
int query2(int a, int num) {
  int ansq = 0;
  for (int i = MAXlg - 1; i >= 0; i--) {
    if (a + (1 << i) < n && dp2[a][i] < num) {
      ansq += (1 << i);
      a += (1 << i);
    }
  }
  return ansq;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> ar[i];
  make_dp1();
  make_dp2();
  if (k == 2) {
    for (int i = 2; i < MAXn; i++)
      ar3[i] = ((long long)ar3[i - 1] + i - 1) % MOD;
  } else {
    ar3[k] = 1;
    ar2[k] = 1;
    for (int i = k + 1; i < MAXn; i++) {
      ar2[i] = ar2[i - 1];
      if (i % (k - 1) == 1) ar2[i]++;
      ar2[i] %= MOD;
      ar3[i] = ((long long)ar3[i - 1] + ar2[i]) % MOD;
      ar3[i] %= MOD;
    }
  }
  for (int i = 0; i < n; i++) {
    long long l = query1(i, ar[i]), r = query2(i, ar[i]),
              hp = ((long long)ar3[l + r + 1] - ar3[l] - ar3[r] + MOD) % MOD,
              hp2 = ar[i], ans2 = (hp * hp2) % MOD;
    ans = ((long long)ans + ans2) % MOD;
  }
  cout << ans;
}

```