#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4.1")
using namespace std;
const int mod = (int)1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int sum(int a, int b) {
  add(a, b);
  return a;
}
int diff(int a, int b) {
  sub(a, b);
  return a;
}
int mult(int a, int b) { return (long long)a * b % mod; }
int n, k;
vector<int> a;
bool read() {
  if (!(cin >> n >> k)) return false;
  a.resize(n);
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  return true;
}
struct segtree {
  int tsz;
  vector<pair<int, int> > vals;
  segtree(const vector<int> &a) {
    tsz = (int)(a).size();
    vals.assign(2 * tsz, make_pair(-1, -1));
    for (int i = 0; i < (int)(tsz); ++i) vals[i + tsz] = make_pair(a[i], i);
    for (int i = ((int)(tsz)-1); i >= 0; --i)
      if (i) vals[i] = max(vals[2 * i], vals[2 * i + 1]);
  }
  pair<int, int> get(int l, int r) const {
    pair<int, int> ans(-1, -1);
    for (l += tsz, r += tsz; l < r; l >>= 1, r >>= 1) {
      if (l & 1) ans = max(ans, vals[l++]);
      if (r & 1) ans = max(ans, vals[--r]);
    }
    return ans;
  }
};
int rec(const int l, const int r, const vector<int> &pref,
        const segtree &data) {
  if (l >= r) return 0;
  const int len = r - l;
  int val, orig_pos;
  tie(val, orig_pos) = data.get(l, r);
  const int pos = orig_pos - l;
  int ans = diff(pref[len], pref[len - pos - 1]);
  sub(ans, pref[pos]);
  ans = mult(ans, val);
  add(ans, rec(l, orig_pos, pref, data));
  add(ans, rec(orig_pos + 1, r, pref, data));
  return ans;
}
int solve() {
  segtree data(a);
  vector<int> pref(n + 1);
  for (int i = 0; i < (int)(n); ++i) pref[i + 1] = sum(pref[i], i / (k - 1));
  int ans = rec(0, n, pref, data);
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  while (read()) {
    auto ans = solve();
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4.1")
using namespace std;
const int mod = (int)1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int sum(int a, int b) {
  add(a, b);
  return a;
}
int diff(int a, int b) {
  sub(a, b);
  return a;
}
int mult(int a, int b) { return (long long)a * b % mod; }
int n, k;
vector<int> a;
bool read() {
  if (!(cin >> n >> k)) return false;
  a.resize(n);
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  return true;
}
struct segtree {
  int tsz;
  vector<pair<int, int> > vals;
  segtree(const vector<int> &a) {
    tsz = (int)(a).size();
    vals.assign(2 * tsz, make_pair(-1, -1));
    for (int i = 0; i < (int)(tsz); ++i) vals[i + tsz] = make_pair(a[i], i);
    for (int i = ((int)(tsz)-1); i >= 0; --i)
      if (i) vals[i] = max(vals[2 * i], vals[2 * i + 1]);
  }
  pair<int, int> get(int l, int r) const {
    pair<int, int> ans(-1, -1);
    for (l += tsz, r += tsz; l < r; l >>= 1, r >>= 1) {
      if (l & 1) ans = max(ans, vals[l++]);
      if (r & 1) ans = max(ans, vals[--r]);
    }
    return ans;
  }
};
int rec(const int l, const int r, const vector<int> &pref,
        const segtree &data) {
  if (l >= r) return 0;
  const int len = r - l;
  int val, orig_pos;
  tie(val, orig_pos) = data.get(l, r);
  const int pos = orig_pos - l;
  int ans = diff(pref[len], pref[len - pos - 1]);
  sub(ans, pref[pos]);
  ans = mult(ans, val);
  add(ans, rec(l, orig_pos, pref, data));
  add(ans, rec(orig_pos + 1, r, pref, data));
  return ans;
}
int solve() {
  segtree data(a);
  vector<int> pref(n + 1);
  for (int i = 0; i < (int)(n); ++i) pref[i + 1] = sum(pref[i], i / (k - 1));
  int ans = rec(0, n, pref, data);
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  while (read()) {
    auto ans = solve();
    cout << ans << endl;
  }
  return 0;
}

```