#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 1000007;
const int MOD = 1000000007;
int n, K;
int a[MAXN], tb[MAXN], b[MAXN], nk;
int stk[MAXN], top;
int q[MAXN];
void init();
void input();
void work();
int add(int x, int y) {
  x += y;
  return (x < MOD) ? x : (x - MOD);
}
void addv(int &x, int y) {
  x += y;
  (x < MOD) ? x : (x -= MOD);
}
int sub(int x, int y) {
  x -= y;
  return (x >= 0) ? x : (x + MOD);
}
void subv(int &x, int y) {
  x -= y;
  (x >= 0) ? x : (x += MOD);
}
int main() {
  init();
  input();
  work();
}
void init() { ios::sync_with_stdio(false); }
void input() {
  scanf("%d%d", &n, &K);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
}
void work() {
  --K;
  int l = 1, r = 0;
  for (int i = n - 1; i >= 0; --i) {
    while (l <= r && a[i] >= a[q[r]]) --r;
    q[++r] = i;
    if (q[l] - i > K) ++l;
    tb[i] = a[q[l]];
  }
  int ans = 0;
  for (int i = 0; i < K; ++i) {
    nk = 0;
    for (int j = i; j < n - K; j += K) b[++nk] = tb[j];
    top = 0;
    stk[0] = 0;
    b[0] = b[nk + 1] = INF;
    for (int j = 1; j <= nk + 1; ++j) {
      while (top && b[j] >= b[stk[top]]) {
        addv(ans, static_cast<long long>(b[stk[top]]) * (j - stk[top]) % MOD *
                      (stk[top] - stk[top - 1]) % MOD);
        --top;
      }
      stk[++top] = j;
    }
  }
  cout << ans << endl;
}

### Prompt

Please formulate 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 = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 1000007;
const int MOD = 1000000007;
int n, K;
int a[MAXN], tb[MAXN], b[MAXN], nk;
int stk[MAXN], top;
int q[MAXN];
void init();
void input();
void work();
int add(int x, int y) {
  x += y;
  return (x < MOD) ? x : (x - MOD);
}
void addv(int &x, int y) {
  x += y;
  (x < MOD) ? x : (x -= MOD);
}
int sub(int x, int y) {
  x -= y;
  return (x >= 0) ? x : (x + MOD);
}
void subv(int &x, int y) {
  x -= y;
  (x >= 0) ? x : (x += MOD);
}
int main() {
  init();
  input();
  work();
}
void init() { ios::sync_with_stdio(false); }
void input() {
  scanf("%d%d", &n, &K);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
}
void work() {
  --K;
  int l = 1, r = 0;
  for (int i = n - 1; i >= 0; --i) {
    while (l <= r && a[i] >= a[q[r]]) --r;
    q[++r] = i;
    if (q[l] - i > K) ++l;
    tb[i] = a[q[l]];
  }
  int ans = 0;
  for (int i = 0; i < K; ++i) {
    nk = 0;
    for (int j = i; j < n - K; j += K) b[++nk] = tb[j];
    top = 0;
    stk[0] = 0;
    b[0] = b[nk + 1] = INF;
    for (int j = 1; j <= nk + 1; ++j) {
      while (top && b[j] >= b[stk[top]]) {
        addv(ans, static_cast<long long>(b[stk[top]]) * (j - stk[top]) % MOD *
                      (stk[top] - stk[top - 1]) % MOD);
        --top;
      }
      stk[++top] = j;
    }
  }
  cout << ans << endl;
}

```