#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000006;
int n, k, a[MAXN], b[MAXN];
int bit[MAXN];
void add(int pos, int val) {
  while (pos < MAXN) {
    bit[pos] = max(bit[pos], val);
    pos += pos & (-pos);
  }
}
int sum(int pos) {
  int re = 0;
  while (pos > 0) {
    re = max(re, bit[pos]);
    pos -= pos & (-pos);
  }
  return re;
}
void init() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = n; i >= 1; i--) {
    add(i, a[i]);
    if (i + k - 1 <= n) {
      b[i] = sum(i + k - 1);
    }
  }
}
struct tup {
  int x, cnt, S;
};
const int MOD = 1e9 + 7;
void push(vector<tup> &sta, int x) {
  int cnt = 1;
  while (sta.size() && sta.back().x <= x) {
    cnt += sta.back().cnt;
    sta.pop_back();
  }
  tup o = {x, cnt, int((long long)x * cnt % MOD)};
  if (sta.size()) {
    o.S += sta.back().S;
    o.S %= MOD;
  }
  sta.push_back(o);
}
vector<tup> sta[MAXN];
int main() {
  init();
  long long ans = 0;
  for (int i = n - k + 1; i >= 1; i--) {
    push(sta[i % (k - 1)], b[i]);
    ans += sta[i % (k - 1)].back().S;
  }
  cout << ans % MOD << '\n';
}

### 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 = 1000006;
int n, k, a[MAXN], b[MAXN];
int bit[MAXN];
void add(int pos, int val) {
  while (pos < MAXN) {
    bit[pos] = max(bit[pos], val);
    pos += pos & (-pos);
  }
}
int sum(int pos) {
  int re = 0;
  while (pos > 0) {
    re = max(re, bit[pos]);
    pos -= pos & (-pos);
  }
  return re;
}
void init() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = n; i >= 1; i--) {
    add(i, a[i]);
    if (i + k - 1 <= n) {
      b[i] = sum(i + k - 1);
    }
  }
}
struct tup {
  int x, cnt, S;
};
const int MOD = 1e9 + 7;
void push(vector<tup> &sta, int x) {
  int cnt = 1;
  while (sta.size() && sta.back().x <= x) {
    cnt += sta.back().cnt;
    sta.pop_back();
  }
  tup o = {x, cnt, int((long long)x * cnt % MOD)};
  if (sta.size()) {
    o.S += sta.back().S;
    o.S %= MOD;
  }
  sta.push_back(o);
}
vector<tup> sta[MAXN];
int main() {
  init();
  long long ans = 0;
  for (int i = n - k + 1; i >= 1; i--) {
    push(sta[i % (k - 1)], b[i]);
    ans += sta[i % (k - 1)].back().S;
  }
  cout << ans % MOD << '\n';
}

```