#include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e6 + 5;
const int MOD = (int)1e9 + 7;
int ls[MAXN], rs[MAXN];
pair<int, int> a[MAXN];
int stk[MAXN], top;
long long ans;
int f[MAXN], g[MAXN], k;
long long cal(int l1, int l2) {
  long long re = 0;
  if (l1 > l2) swap(l1, l2);
  for (int i = 0; i <= l1; i++) {
    re += (i + l2) / (k - 1) - max(0, i - 1) / (k - 1);
  }
  re %= MOD;
  return re;
}
void dfs(int u, int l, int r) {
  ans = (ans + cal(a[u].second - l, r - a[u].second) * -a[u].first) % MOD;
  if (ls[u]) dfs(ls[u], l, a[u].second - 1);
  if (rs[u]) dfs(rs[u], a[u].second + 1, r);
}
int main() {
  int n, rt = 1;
  scanf("%d%d", &n, &k);
  int now = k;
  for (int i = 1;; now += k - 1, i++) {
    now = min(now, n);
    f[now] = i;
    if (now == n) break;
  }
  for (int i = 1; i <= n; i++)
    if (!f[i]) f[i] = f[i - 1];
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = make_pair(-x, i);
    ls[i] = rs[i] = 0;
    if (a[i] < a[rt]) rt = i;
  }
  top = 0;
  for (int i = 1; i <= n; i++) {
    int tmp = top;
    while (tmp && a[stk[tmp - 1]] > a[i]) tmp--;
    if (tmp) rs[stk[tmp - 1]] = i;
    if (top > tmp) ls[i] = stk[tmp];
    stk[tmp++] = i;
    top = tmp;
  }
  dfs(rt, 1, n);
  printf("%lld\n", ans);
  return 0;
}

### 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 = (int)1e6 + 5;
const int MOD = (int)1e9 + 7;
int ls[MAXN], rs[MAXN];
pair<int, int> a[MAXN];
int stk[MAXN], top;
long long ans;
int f[MAXN], g[MAXN], k;
long long cal(int l1, int l2) {
  long long re = 0;
  if (l1 > l2) swap(l1, l2);
  for (int i = 0; i <= l1; i++) {
    re += (i + l2) / (k - 1) - max(0, i - 1) / (k - 1);
  }
  re %= MOD;
  return re;
}
void dfs(int u, int l, int r) {
  ans = (ans + cal(a[u].second - l, r - a[u].second) * -a[u].first) % MOD;
  if (ls[u]) dfs(ls[u], l, a[u].second - 1);
  if (rs[u]) dfs(rs[u], a[u].second + 1, r);
}
int main() {
  int n, rt = 1;
  scanf("%d%d", &n, &k);
  int now = k;
  for (int i = 1;; now += k - 1, i++) {
    now = min(now, n);
    f[now] = i;
    if (now == n) break;
  }
  for (int i = 1; i <= n; i++)
    if (!f[i]) f[i] = f[i - 1];
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = make_pair(-x, i);
    ls[i] = rs[i] = 0;
    if (a[i] < a[rt]) rt = i;
  }
  top = 0;
  for (int i = 1; i <= n; i++) {
    int tmp = top;
    while (tmp && a[stk[tmp - 1]] > a[i]) tmp--;
    if (tmp) rs[stk[tmp - 1]] = i;
    if (top > tmp) ls[i] = stk[tmp];
    stk[tmp++] = i;
    top = tmp;
  }
  dfs(rt, 1, n);
  printf("%lld\n", ans);
  return 0;
}

```