#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e6 + 7;
const long long MOD = 1000 * 1000 * 1000 + 7;
long long n, k, len;
long long a[MAXN];
void read() {
  cin >> n >> k;
  len = k - 1;
  for (long long i = 0; i < n; ++i) cin >> a[i];
}
long long l[MAXN], r[MAXN];
void precalc() {
  for (long long i = 0; i < n; ++i) {
    l[i] = -1;
    r[i] = n;
  }
  vector<long long> st;
  for (long long i = 0; i < n; ++i) {
    while (st.size() && a[st.back()] < a[i]) {
      r[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  st.clear();
  for (long long i = n - 1; i >= 0; --i) {
    while (st.size() && a[st.back()] <= a[i]) {
      l[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
}
long long get(long long a, long long b) {
  long long ta = (a + 1) / len;
  a = (a + 1) % len;
  long long tb = (b + 1) / len;
  b = (b + 1) % len;
  long long ans = 0;
  ans += ta * tb * len;
  ans += a * tb;
  ans += b * ta;
  ans += (a && b);
  long long tl = max(1ll, len - b + 1);
  long long tr = a - 1;
  ans += max(0ll, tr - tl + 1);
  ans %= MOD;
  return ans;
}
long long ans = 0;
void solve() {
  precalc();
  for (long long i = 0; i < n; ++i) {
    long long llen = i - l[i] - 1;
    long long rlen = r[i] - 1 - i;
    long long f = get(llen, rlen) - 1;
    ans += a[i] * f;
    ans = (ans % MOD + MOD) % MOD;
  }
}
void print() { cout << ans << '\n'; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
  print();
  return 0;
}

### Prompt

Your challenge is to write 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 long long MAXN = 1e6 + 7;
const long long MOD = 1000 * 1000 * 1000 + 7;
long long n, k, len;
long long a[MAXN];
void read() {
  cin >> n >> k;
  len = k - 1;
  for (long long i = 0; i < n; ++i) cin >> a[i];
}
long long l[MAXN], r[MAXN];
void precalc() {
  for (long long i = 0; i < n; ++i) {
    l[i] = -1;
    r[i] = n;
  }
  vector<long long> st;
  for (long long i = 0; i < n; ++i) {
    while (st.size() && a[st.back()] < a[i]) {
      r[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  st.clear();
  for (long long i = n - 1; i >= 0; --i) {
    while (st.size() && a[st.back()] <= a[i]) {
      l[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
}
long long get(long long a, long long b) {
  long long ta = (a + 1) / len;
  a = (a + 1) % len;
  long long tb = (b + 1) / len;
  b = (b + 1) % len;
  long long ans = 0;
  ans += ta * tb * len;
  ans += a * tb;
  ans += b * ta;
  ans += (a && b);
  long long tl = max(1ll, len - b + 1);
  long long tr = a - 1;
  ans += max(0ll, tr - tl + 1);
  ans %= MOD;
  return ans;
}
long long ans = 0;
void solve() {
  precalc();
  for (long long i = 0; i < n; ++i) {
    long long llen = i - l[i] - 1;
    long long rlen = r[i] - 1 - i;
    long long f = get(llen, rlen) - 1;
    ans += a[i] * f;
    ans = (ans % MOD + MOD) % MOD;
  }
}
void print() { cout << ans << '\n'; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
  print();
  return 0;
}

```