#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
const int maxn = 1000005;
const int off = 1 << 20;
const int mod = 1e9 + 7;
int cntk[maxn];
int ways[maxn];
pii tree[2 * off];
pii get(int l, int r) {
  pii ret;
  for (l += off, r += off; l < r; l /= 2, r /= 2) {
    if (l & 1) ret = max(ret, tree[l++]);
    if (r & 1) ret = max(ret, tree[--r]);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0, a; i < n; ++i) cin >> a, tree[i + off] = {a, i};
  for (int i = off - 1; i; --i) tree[i] = max(tree[2 * i], tree[2 * i + 1]);
  for (int i = k; i <= n; i += k - 1) cntk[i]++;
  for (int i = 1; i <= n; ++i) {
    cntk[i] += cntk[i - 1];
    ways[i] += ways[i - 1] + cntk[i];
    if (ways[i] >= mod) ways[i] -= mod;
  }
  stack<tuple<int, int, int>> stk;
  stk.emplace(0, n, 0);
  long long sol = 0;
  while (!stk.empty()) {
    int l, r, pv, cv, i;
    tie(l, r, pv) = stk.top();
    stk.pop();
    if (r - l < k) continue;
    tie(cv, i) = get(l, r);
    sol += 1LL * (cv - pv) * ways[r - l] % mod;
    stk.emplace(l, i, cv);
    stk.emplace(i + 1, r, cv);
  }
  cout << (sol % mod + mod) % mod;
}

### Prompt

In Cpp, your task is to solve 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;
using pii = pair<int, int>;
const int maxn = 1000005;
const int off = 1 << 20;
const int mod = 1e9 + 7;
int cntk[maxn];
int ways[maxn];
pii tree[2 * off];
pii get(int l, int r) {
  pii ret;
  for (l += off, r += off; l < r; l /= 2, r /= 2) {
    if (l & 1) ret = max(ret, tree[l++]);
    if (r & 1) ret = max(ret, tree[--r]);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0, a; i < n; ++i) cin >> a, tree[i + off] = {a, i};
  for (int i = off - 1; i; --i) tree[i] = max(tree[2 * i], tree[2 * i + 1]);
  for (int i = k; i <= n; i += k - 1) cntk[i]++;
  for (int i = 1; i <= n; ++i) {
    cntk[i] += cntk[i - 1];
    ways[i] += ways[i - 1] + cntk[i];
    if (ways[i] >= mod) ways[i] -= mod;
  }
  stack<tuple<int, int, int>> stk;
  stk.emplace(0, n, 0);
  long long sol = 0;
  while (!stk.empty()) {
    int l, r, pv, cv, i;
    tie(l, r, pv) = stk.top();
    stk.pop();
    if (r - l < k) continue;
    tie(cv, i) = get(l, r);
    sol += 1LL * (cv - pv) * ways[r - l] % mod;
    stk.emplace(l, i, cv);
    stk.emplace(i + 1, r, cv);
  }
  cout << (sol % mod + mod) % mod;
}

```