#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long INF = 1e18L + 7;
template <class n, class second>
ostream &operator<<(ostream &p, pair<n, second> x) {
  return p << "<" << x.first << ", " << x.second << ">";
}
template <class n>
auto operator<<(ostream &p, n y) ->
    typename enable_if<!is_same<n, string>::value,
                       decltype(y.begin(), p)>::type {
  int o = 0;
  p << "{";
  for (auto c : y) {
    if (o++) p << ", ";
    p << c;
  }
  return p << "}";
}
void dor() { cerr << '\n'; }
template <class n, class... second>
void dor(n p, second... y) {
  cerr << p << " ";
  dor(y...);
}
template <class n, class second>
void mini(n &p, second y) {
  if (p > y) p = y;
}
template <class n, class second>
void maxi(n &p, second y) {
  if (p < y) p = y;
}
const long long mod = 1e9 + 7;
const long long N = 1e6 + 7;
long long n, k, ans;
long long a[N];
long long lewo[N];
long long prawo[N];
long long ile_rund(long long x) {
  long long l = 0, r = x;
  while (l < r) {
    long long m = (l + r) / 2;
    if (l == m) ++m;
    if (x - (m - 1) * (k - 1) >= k)
      l = m;
    else
      r = m - 1;
  }
  return l;
}
long long zera(long long x) {
  long long l = 0, r = x;
  while (l < r) {
    long long m = (l + r) / 2;
    if (l == m) ++m;
    if (x - m * (k - 1) >= 0)
      l = m;
    else
      r = m - 1;
  }
  return l;
}
long long wklad(long long l, long long p) {
  long long x = 1 + l + p;
  long long rund = ile_rund(x);
  if (!rund) return 0;
  ;
  long long raz = x - (k - 1);
  long long dwa = x - rund * (k - 1);
  long long len = rund * (raz + dwa) / 2;
  ;
  long long rund_lewo = zera(l);
  long long rund_prawo = zera(p);
  l -= k - 1;
  p -= k - 1;
  long long zer = rund_lewo * (2 * l - (rund_lewo - 1) * (k - 1)) / 2 +
                  rund_prawo * (2 * p - (rund_prawo - 1) * (k - 1)) / 2;
  return len - zer;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> k;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  a[0] = a[n + 1] = inf;
  lewo[1] = 0;
  for (long long i = 2; i <= n; ++i) {
    long long j = i - 1;
    while (make_pair(a[j], j) < make_pair(a[i], i)) j = lewo[j];
    lewo[i] = j;
  }
  prawo[n] = n + 1;
  for (long long i = n - 1; i >= 1; --i) {
    long long j = i + 1;
    while (make_pair(a[j], j) < make_pair(a[i], i)) j = prawo[j];
    prawo[i] = j;
  }
  for (long long i = 1; i <= n; ++i) {
    long long l = i - lewo[i] - 1;
    long long p = prawo[i] - i - 1;
    ans += wklad(l, p) % mod * a[i] % mod;
    ans %= mod;
  }
  ans %= mod;
  ans += mod;
  ans %= mod;
  cout << ans << '\n';
}

### 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;
const int inf = 1e9 + 7;
const long long INF = 1e18L + 7;
template <class n, class second>
ostream &operator<<(ostream &p, pair<n, second> x) {
  return p << "<" << x.first << ", " << x.second << ">";
}
template <class n>
auto operator<<(ostream &p, n y) ->
    typename enable_if<!is_same<n, string>::value,
                       decltype(y.begin(), p)>::type {
  int o = 0;
  p << "{";
  for (auto c : y) {
    if (o++) p << ", ";
    p << c;
  }
  return p << "}";
}
void dor() { cerr << '\n'; }
template <class n, class... second>
void dor(n p, second... y) {
  cerr << p << " ";
  dor(y...);
}
template <class n, class second>
void mini(n &p, second y) {
  if (p > y) p = y;
}
template <class n, class second>
void maxi(n &p, second y) {
  if (p < y) p = y;
}
const long long mod = 1e9 + 7;
const long long N = 1e6 + 7;
long long n, k, ans;
long long a[N];
long long lewo[N];
long long prawo[N];
long long ile_rund(long long x) {
  long long l = 0, r = x;
  while (l < r) {
    long long m = (l + r) / 2;
    if (l == m) ++m;
    if (x - (m - 1) * (k - 1) >= k)
      l = m;
    else
      r = m - 1;
  }
  return l;
}
long long zera(long long x) {
  long long l = 0, r = x;
  while (l < r) {
    long long m = (l + r) / 2;
    if (l == m) ++m;
    if (x - m * (k - 1) >= 0)
      l = m;
    else
      r = m - 1;
  }
  return l;
}
long long wklad(long long l, long long p) {
  long long x = 1 + l + p;
  long long rund = ile_rund(x);
  if (!rund) return 0;
  ;
  long long raz = x - (k - 1);
  long long dwa = x - rund * (k - 1);
  long long len = rund * (raz + dwa) / 2;
  ;
  long long rund_lewo = zera(l);
  long long rund_prawo = zera(p);
  l -= k - 1;
  p -= k - 1;
  long long zer = rund_lewo * (2 * l - (rund_lewo - 1) * (k - 1)) / 2 +
                  rund_prawo * (2 * p - (rund_prawo - 1) * (k - 1)) / 2;
  return len - zer;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> k;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  a[0] = a[n + 1] = inf;
  lewo[1] = 0;
  for (long long i = 2; i <= n; ++i) {
    long long j = i - 1;
    while (make_pair(a[j], j) < make_pair(a[i], i)) j = lewo[j];
    lewo[i] = j;
  }
  prawo[n] = n + 1;
  for (long long i = n - 1; i >= 1; --i) {
    long long j = i + 1;
    while (make_pair(a[j], j) < make_pair(a[i], i)) j = prawo[j];
    prawo[i] = j;
  }
  for (long long i = 1; i <= n; ++i) {
    long long l = i - lewo[i] - 1;
    long long p = prawo[i] - i - 1;
    ans += wklad(l, p) % mod * a[i] % mod;
    ans %= mod;
  }
  ans %= mod;
  ans += mod;
  ans %= mod;
  cout << ans << '\n';
}

```