#include <bits/stdc++.h>
using namespace std;
template <class V, int NV>
class SegTree_1 {
 public:
  vector<V> val;
  static V const def = 0;
  V comp(V l, V r) { return max(l, r); };
  SegTree_1() { val = vector<V>(NV * 2, def); };
  V getval(int x, int y, int l = 0, int r = NV, int k = 1) {
    if (r <= x || y <= l) return def;
    if (x <= l && r <= y) return val[k];
    return comp(getval(x, y, l, (l + r) / 2, k * 2),
                getval(x, y, (l + r) / 2, r, k * 2 + 1));
  }
  void update(int entry, V v) {
    entry += NV;
    val[entry] = comp(v, val[entry]);
    while (entry > 1)
      entry >>= 1, val[entry] = comp(val[entry * 2], val[entry * 2 + 1]);
  }
};
SegTree_1<signed long long, 1 << 21> st;
int N, K;
int A[1010101];
vector<int> C[1010101];
signed long long mo = 1000000007;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  scanf("%d%d", &N, &K);
  for (i = 0; i < (N); i++) {
    scanf("%d", &A[i]);
    st.update(i, A[i]);
    if (i + K <= N) C[i % (K - 1)].push_back(i);
  }
  signed long long ret = 0;
  for (i = 0; i < (K); i++) {
    map<signed long long, signed long long> M;
    signed long long tot = 0;
    for (auto& c : C[i]) {
      signed long long ma = st.getval(c, c + K);
      (tot += ma) %= mo;
      M[ma]++;
      while (M.begin()->first < ma) {
        (tot += (ma - M.begin()->first) * M.begin()->second) %= mo;
        M[ma] += M.begin()->second;
        M.erase(M.begin());
      }
      (ret += tot) %= mo;
    }
  }
  cout << ret << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

### 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;
template <class V, int NV>
class SegTree_1 {
 public:
  vector<V> val;
  static V const def = 0;
  V comp(V l, V r) { return max(l, r); };
  SegTree_1() { val = vector<V>(NV * 2, def); };
  V getval(int x, int y, int l = 0, int r = NV, int k = 1) {
    if (r <= x || y <= l) return def;
    if (x <= l && r <= y) return val[k];
    return comp(getval(x, y, l, (l + r) / 2, k * 2),
                getval(x, y, (l + r) / 2, r, k * 2 + 1));
  }
  void update(int entry, V v) {
    entry += NV;
    val[entry] = comp(v, val[entry]);
    while (entry > 1)
      entry >>= 1, val[entry] = comp(val[entry * 2], val[entry * 2 + 1]);
  }
};
SegTree_1<signed long long, 1 << 21> st;
int N, K;
int A[1010101];
vector<int> C[1010101];
signed long long mo = 1000000007;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  scanf("%d%d", &N, &K);
  for (i = 0; i < (N); i++) {
    scanf("%d", &A[i]);
    st.update(i, A[i]);
    if (i + K <= N) C[i % (K - 1)].push_back(i);
  }
  signed long long ret = 0;
  for (i = 0; i < (K); i++) {
    map<signed long long, signed long long> M;
    signed long long tot = 0;
    for (auto& c : C[i]) {
      signed long long ma = st.getval(c, c + K);
      (tot += ma) %= mo;
      M[ma]++;
      while (M.begin()->first < ma) {
        (tot += (ma - M.begin()->first) * M.begin()->second) %= mo;
        M[ma] += M.begin()->second;
        M.erase(M.begin());
      }
      (ret += tot) %= mo;
    }
  }
  cout << ret << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

```